message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A coder cannot sit and code all day. Sometimes it is a good idea to rise from the desk, have a rest, have small talk with colleagues and even play. The coders of the F company have their favorite ball game.
Let's imagine the game on the plane with a cartesian coordinate system. The point (0, 0) contains the player who chooses an arbitrary direction and throws a ball in that direction. The ball hits the plane at distance d from the player's original position and continues flying in the same direction. After the ball hits the plane for the first time, it flies on and hits the plane again at distance 2·d from the player's original position and so on (it continue flying in the chosen direction and hitting the plane after each d units). All coders in the F company are strong, so the ball flies infinitely far away.
The plane has n circles painted on it. If a ball hits the plane and hits a circle that is painted on the plane (including its border), then the player gets one point. The ball can hit multiple circles at once and get one point for each of them (if the ball hits some circle x times during the move, the player also gets x points). Count the maximum number of points a player can get if he throws a ball in the arbitrary direction. Note that the direction may have real cooridinates.
Input
The first line contains two space-separated integers — n и d (1 ≤ n ≤ 2·104; 5 ≤ d ≤ 10). Next n lines contain the circles' description. The i-th line contains three space-separated integers xi, yi, ri ( - 10000 ≤ xi, yi ≤ 10000; 1 ≤ r ≤ 50), where (xi, yi, ri) are the coordinates of the center and the radius of the circle, correspondingly. The point (0, 0) is not inside or on the border of some circle.
Output
Print a single integer — the maximum number of points you can get.
Examples
Input
2 5
1 1 1
5 0 1
Output
1
Input
2 5
4 0 3
5 3 1
Output
2
Input
1 10
20 0 10
Output
3
Submitted Solution:
```
import math
class Point:
def __init__(self, x1, y1):
self.x = x1
self.y = y1
class Circle:
def __init__(self, x1, y1, r1):
self.A = Point(0, 0)
self.B = Point(0, 0)
self.x = float(x1)
self.y = float(y1)
self.r = float(r1)
self.A.x = self.x - self.r #A is bottom left vertex
self.A.y = self.y - self.r
self.B.x = self.x + self.r #B is top right vertex
self.B.y = self.y + self.r
self.angle = math.degrees(math.atan2(self.y, self.x))
#special class that has 5 lists to store circles in
class SpecialCircleList:
def append(self, c):
if(c.x >= 0):
if(c.x - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.x <= 0):
if(c.x + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.y >= 0):
if(c.y - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.y <= 0):
if(c.y + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.x > 0 and c.y > 0):
self.first_quadrant.append(c)
elif(c.x < 0 and c.y > 0):
self.second_quadrant.append(c)
elif(c.x < 0 and c.y < 0):
self.thrid_quadrant.append(c)
else:
self.fourth_quadrant.append(c)
#pass in a bounce point or unit vector, and i'll return a list of circles to check
def getRelevantList(self, p):
acceptance_angle = 45
pangle = math.degrees(math.atan2(p.y, p.x))
retList = list()
for c in self.crosses_axis:
if((abs(pangle - c.angle) <acceptance_angle) or abs(c.x < 50) or abs(c.y<50)):
retList.append(c)
#retList.extend(self.crosses_axis)
if(p.x >= 0 and p.y >= 0):
#retList.extend(self.first_quadrant)
for c in self.first_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
elif(p.x < 0 and p.y >= 0):
#retList.extend(self.second_quadrant)
for c in self.second_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
elif(p.x < 0 and p.y < 0):
#retList.extend(self.thrid_quadrant)
for c in self.thrid_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
else:
#retList.extend(self.fourth_quadrant)
for c in self.fourth_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
return(retList)
def __init__(self):
self.crosses_axis = list() #for any circle that even remotely crosses and axis
self.first_quadrant = list() #for any circle that exclusively resides in quad
self.second_quadrant = list()
self.thrid_quadrant = list()
self.fourth_quadrant = list()
circle_list = SpecialCircleList()
d = 0
bound_up = 0
bound_down = 0
bound_left = 0
bound_right = 0
max_score = 0
def main():
global circle_list
global max_score
global d
global bound_up
global bound_down
global bound_left
global bound_right
line_in = input("")
(n, d) = line_in.split()
n = int(n)
d = int(d)
for i in range(n):
line_in = input("")
(x, y, r) = line_in.split()
x = int(x)
y = int(y)
r = int(r)
newCircle = Circle(x, y, r)
circle_list.append(newCircle)
#determine checking bounds
if(y+r > bound_up):
bound_up = y + r
if(y-r < bound_down):
bound_down = y - r
if(x + r > bound_right):
bound_right = x + r
if(x - r < bound_left):
bound_left = x - r
#brute force for angles 0->360
for angle in range(0, 360, 3):
#generate unit vector with direction
unit_vector = Point(math.cos(angle*math.pi/180),math.sin(angle*math.pi/180))
#get a list of points where the ball will bounce for a given direction/line
bounce_list = getBouncePoints(unit_vector, bound_up, bound_right,\
bound_down, bound_left)
if(len(bounce_list) <= max_score):
continue #no point doing this one if i've already got a higher score
score_for_line = checkIfInAnyCircles(bounce_list, unit_vector)
if(score_for_line > max_score):
max_score = score_for_line
print(max_score)
#returns the score of a line/trajectory
def checkIfInAnyCircles(bounce_list, unit_vector):
global circle_list
list_of_relavent_circles = circle_list.getRelevantList(unit_vector)
score = 0
if(len(list_of_relavent_circles) == 0 or len(bounce_list) == 0):
return(0)
for b in bounce_list:
for c in list_of_relavent_circles:
if(b.x >= c.A.x and b.y >= c.A.y and b.x <= c.B.x and b.y <= c.B.y):
if(thoroughCheck(b, c)):
score += 1
return(score)
#after passing a basic bounded box collision check, do a more thorough check with
#floating point arithmatic to see if a bounce (b) lands within a given circle, (c)
def thoroughCheck(b, c):
distanceFromCircleCentre = (b.x - c.x)**2 + (b.y - c.y)**2
distanceFromCircleCentre = math.sqrt(distanceFromCircleCentre)
if(distanceFromCircleCentre <= c.r):
return(True)
else:
return(False)
#returns a list of points which a ball will bounce at given a direction/trajectory
#takes into consideration bounding box, so doens't calculate more than necessary
def getBouncePoints(unit_vector, bound_up, bound_right, bound_down, bound_left ):
global d
bounce_list = list()
bounce_vector = Point(unit_vector.x * d, unit_vector.y * d)
newBounce = Point(0, 0)
if(unit_vector.x >= 0 and unit_vector.y >= 0):
#first quadrant
while(newBounce.x <= bound_right and newBounce.y <= bound_up):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
elif(unit_vector.x < 0 and unit_vector.y >= 0):
#second quadrant
while(newBounce.x >= bound_left and newBounce.y <= bound_up):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
elif(unit_vector.x < 0 and unit_vector.y < 0):
#thrid quadrant
while(newBounce.x >= bound_left and newBounce.y >= bound_down):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
else:
#fourth quadrant
while(newBounce.x <= bound_left and newBounce.y >= bound_down):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
return(bounce_list)
if __name__ == "__main__":
main()
``` | instruction | 0 | 6,169 | 3 | 12,338 |
No | output | 1 | 6,169 | 3 | 12,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A coder cannot sit and code all day. Sometimes it is a good idea to rise from the desk, have a rest, have small talk with colleagues and even play. The coders of the F company have their favorite ball game.
Let's imagine the game on the plane with a cartesian coordinate system. The point (0, 0) contains the player who chooses an arbitrary direction and throws a ball in that direction. The ball hits the plane at distance d from the player's original position and continues flying in the same direction. After the ball hits the plane for the first time, it flies on and hits the plane again at distance 2·d from the player's original position and so on (it continue flying in the chosen direction and hitting the plane after each d units). All coders in the F company are strong, so the ball flies infinitely far away.
The plane has n circles painted on it. If a ball hits the plane and hits a circle that is painted on the plane (including its border), then the player gets one point. The ball can hit multiple circles at once and get one point for each of them (if the ball hits some circle x times during the move, the player also gets x points). Count the maximum number of points a player can get if he throws a ball in the arbitrary direction. Note that the direction may have real cooridinates.
Input
The first line contains two space-separated integers — n и d (1 ≤ n ≤ 2·104; 5 ≤ d ≤ 10). Next n lines contain the circles' description. The i-th line contains three space-separated integers xi, yi, ri ( - 10000 ≤ xi, yi ≤ 10000; 1 ≤ r ≤ 50), where (xi, yi, ri) are the coordinates of the center and the radius of the circle, correspondingly. The point (0, 0) is not inside or on the border of some circle.
Output
Print a single integer — the maximum number of points you can get.
Examples
Input
2 5
1 1 1
5 0 1
Output
1
Input
2 5
4 0 3
5 3 1
Output
2
Input
1 10
20 0 10
Output
3
Submitted Solution:
```
import math
class Point:
def __init__(self, x1, y1):
self.x = x1
self.y = y1
class Circle:
def __init__(self, x1, y1, r1):
self.A = Point(0, 0)
self.B = Point(0, 0)
self.x = float(x1)
self.y = float(y1)
self.r = float(r1)
self.A.x = self.x - self.r #A is bottom left vertex
self.A.y = self.y - self.r
self.B.x = self.x + self.r #B is top right vertex
self.B.y = self.y + self.r
self.angle = math.degrees(math.atan2(self.y, self.x))
self.d = math.hypot(self.x, self.y)
#special class that has 5 lists to store circles in
class SpecialCircleList:
def append(self, c):
if(c.x >= 0):
if(c.x - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.x <= 0):
if(c.x + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.y >= 0):
if(c.y - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.y <= 0):
if(c.y + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.x > 0 and c.y > 0):
self.first_quadrant.append(c)
elif(c.x < 0 and c.y > 0):
self.second_quadrant.append(c)
elif(c.x < 0 and c.y < 0):
self.thrid_quadrant.append(c)
else:
self.fourth_quadrant.append(c)
#pass in a bounce point or unit vector, and i'll return a list of circles to check
def getRelevantList(self, p):
acceptance_angle = 30
pangle = math.degrees(math.atan2(p.y, p.x))
retList = list()
max_d = 0
for c in self.crosses_axis:
if((abs(pangle - c.angle) <acceptance_angle) or abs(c.x < 50) or abs(c.y<50)):
retList.append(c)
if(c.d > max_d):
max_d = c.d
#retList.extend(self.crosses_axis)
if(p.x >= 0 and p.y >= 0):
#retList.extend(self.first_quadrant)
for c in self.first_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
elif(p.x < 0 and p.y >= 0):
#retList.extend(self.second_quadrant)
for c in self.second_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
elif(p.x < 0 and p.y < 0):
#retList.extend(self.thrid_quadrant)
for c in self.thrid_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
else:
#retList.extend(self.fourth_quadrant)
for c in self.fourth_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
return(retList, max_d)
def __init__(self):
self.crosses_axis = list() #for any circle that even remotely crosses and axis
self.first_quadrant = list() #for any circle that exclusively resides in quad
self.second_quadrant = list()
self.thrid_quadrant = list()
self.fourth_quadrant = list()
circle_list = SpecialCircleList()
d = 0
bound_up = 0
bound_down = 0
bound_left = 0
bound_right = 0
max_score = 0
def main():
global circle_list
global max_score
global d
global bound_up
global bound_down
global bound_left
global bound_right
line_in = input("")
(n, d) = line_in.split()
n = int(n)
d = int(d)
for i in range(n):
line_in = input("")
(x, y, r) = line_in.split()
x = int(x)
y = int(y)
r = int(r)
newCircle = Circle(x, y, r)
circle_list.append(newCircle)
#determine checking bounds
if(y+r > bound_up):
bound_up = y + r
if(y-r < bound_down):
bound_down = y - r
if(x + r > bound_right):
bound_right = x + r
if(x - r < bound_left):
bound_left = x - r
#brute force for angles 0->360
for angle in range(0, 360, 2):
#generate unit vector with direction
unit_vector = Point(math.cos(angle*math.pi/180),math.sin(angle*math.pi/180))
#get a list of points where the ball will bounce for a given direction/line
score_for_line = checkIfInAnyCircles(unit_vector)
if(score_for_line > max_score):
max_score = score_for_line
print(max_score)
#returns the score of a line/trajectory
def checkIfInAnyCircles(unit_vector):
global circle_list
global d
list_of_relavent_circles, max_d = circle_list.getRelevantList(unit_vector)
#bounce_list = getBouncePoints(unit_vector, bound_up, bound_right,\
# bound_down, bound_left)
score = 0
if(len(list_of_relavent_circles) < (max_score/2)):
return(0)
#largestX = 0
#largestY = 0
#for c in list_of_relavent_circles:
# if(abs(c.x) > largestX):
# largestX = abs(c.x)
# if(abs(c.y) > largestY):
# largestY = abs(c.y)
bounce_vector = Point(unit_vector.x * d, unit_vector.y * d)
b = Point(bounce_vector.x, bounce_vector.y)
max_d = int((max_d + 50)/d)
i = 0
while(i < max_d):
#while((abs(b.x) <= largestX + 50) and (abs(b.y) <= largestY + 50)):
#for b in bounce_list:
for c in list_of_relavent_circles:
if(b.x >= c.A.x and b.y >= c.A.y and b.x <= c.B.x and b.y <= c.B.y):
if(thoroughCheck(b, c)):
score += 1
b.x += bounce_vector.x
b.y += bounce_vector.y
i += 1
return(score)
#after passing a basic bounded box collision check, do a more thorough check with
#floating point arithmatic to see if a bounce (b) lands within a given circle, (c)
def thoroughCheck(b, c):
distanceFromCircleCentre = (b.x - c.x)**2 + (b.y - c.y)**2
distanceFromCircleCentre = math.sqrt(distanceFromCircleCentre)
if(distanceFromCircleCentre <= c.r):
return(True)
else:
return(False)
#returns a list of points which a ball will bounce at given a direction/trajectory
#takes into consideration bounding box, so doens't calculate more than necessary
def getBouncePoints(unit_vector, bound_up, bound_right, bound_down, bound_left ):
global d
bounce_list = list()
bounce_vector = Point(unit_vector.x * d, unit_vector.y * d)
newBounce = Point(0, 0)
if(unit_vector.x >= 0 and unit_vector.y >= 0):
#first quadrant
while(newBounce.x <= bound_right and newBounce.y <= bound_up):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
elif(unit_vector.x < 0 and unit_vector.y >= 0):
#second quadrant
while(newBounce.x >= bound_left and newBounce.y <= bound_up):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
elif(unit_vector.x < 0 and unit_vector.y < 0):
#thrid quadrant
while(newBounce.x >= bound_left and newBounce.y >= bound_down):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
else:
#fourth quadrant
while(newBounce.x <= bound_left and newBounce.y >= bound_down):
newBounce = Point(newBounce.x +bounce_vector.x, newBounce.y +bounce_vector.y)
bounce_list.append(newBounce)
return(bounce_list)
if __name__ == "__main__":
main()
``` | instruction | 0 | 6,170 | 3 | 12,340 |
No | output | 1 | 6,170 | 3 | 12,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A coder cannot sit and code all day. Sometimes it is a good idea to rise from the desk, have a rest, have small talk with colleagues and even play. The coders of the F company have their favorite ball game.
Let's imagine the game on the plane with a cartesian coordinate system. The point (0, 0) contains the player who chooses an arbitrary direction and throws a ball in that direction. The ball hits the plane at distance d from the player's original position and continues flying in the same direction. After the ball hits the plane for the first time, it flies on and hits the plane again at distance 2·d from the player's original position and so on (it continue flying in the chosen direction and hitting the plane after each d units). All coders in the F company are strong, so the ball flies infinitely far away.
The plane has n circles painted on it. If a ball hits the plane and hits a circle that is painted on the plane (including its border), then the player gets one point. The ball can hit multiple circles at once and get one point for each of them (if the ball hits some circle x times during the move, the player also gets x points). Count the maximum number of points a player can get if he throws a ball in the arbitrary direction. Note that the direction may have real cooridinates.
Input
The first line contains two space-separated integers — n и d (1 ≤ n ≤ 2·104; 5 ≤ d ≤ 10). Next n lines contain the circles' description. The i-th line contains three space-separated integers xi, yi, ri ( - 10000 ≤ xi, yi ≤ 10000; 1 ≤ r ≤ 50), where (xi, yi, ri) are the coordinates of the center and the radius of the circle, correspondingly. The point (0, 0) is not inside or on the border of some circle.
Output
Print a single integer — the maximum number of points you can get.
Examples
Input
2 5
1 1 1
5 0 1
Output
1
Input
2 5
4 0 3
5 3 1
Output
2
Input
1 10
20 0 10
Output
3
Submitted Solution:
```
import math
class Point:
def __init__(self, x1, y1):
self.x = x1
self.y = y1
class Circle:
def __init__(self, x1, y1, r1):
self.A = Point(0, 0)
self.B = Point(0, 0)
self.x = float(x1)
self.y = float(y1)
self.r = float(r1)
self.A.x = self.x - self.r #A is bottom left vertex
self.A.y = self.y - self.r
self.B.x = self.x + self.r #B is top right vertex
self.B.y = self.y + self.r
self.angle = math.degrees(math.atan2(self.y, self.x))
self.d = math.hypot(self.x, self.y)
#special class that has 5 lists to store circles in
class SpecialCircleList:
def append(self, c):
if(c.x >= 0):
if(c.x - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.x <= 0):
if(c.x + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.y >= 0):
if(c.y - c.r <= 0):
self.crosses_axis.append(c)
return
if(c.y <= 0):
if(c.y + c.r >= 0):
self.crosses_axis.append(c)
return
if(c.x > 0 and c.y > 0):
self.first_quadrant.append(c)
elif(c.x < 0 and c.y > 0):
self.second_quadrant.append(c)
elif(c.x < 0 and c.y < 0):
self.thrid_quadrant.append(c)
else:
self.fourth_quadrant.append(c)
#pass in a bounce point or unit vector, and i'll return a list of circles to check
def getRelevantList(self, p):
acceptance_angle = 5
pangle = math.degrees(math.atan2(p.y, p.x))
retList = list()
max_d = 0
for c in self.crosses_axis:
if((abs(pangle - c.angle) <acceptance_angle) or abs(c.x < 50) or abs(c.y<50)):
retList.append(c)
if(c.d > max_d):
max_d = c.d
#retList.extend(self.crosses_axis)
if(p.x >= 0 and p.y >= 0):
#retList.extend(self.first_quadrant)
for c in self.first_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
elif(p.x < 0 and p.y >= 0):
#retList.extend(self.second_quadrant)
for c in self.second_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
elif(p.x < 0 and p.y < 0):
#retList.extend(self.thrid_quadrant)
for c in self.thrid_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
else:
#retList.extend(self.fourth_quadrant)
for c in self.fourth_quadrant:
if(abs(pangle - c.angle) < acceptance_angle):
retList.append(c)
if(c.d > max_d):
max_d = c.d
return(retList, max_d)
def __init__(self):
self.crosses_axis = list() #for any circle that even remotely crosses and axis
self.first_quadrant = list() #for any circle that exclusively resides in quad
self.second_quadrant = list()
self.thrid_quadrant = list()
self.fourth_quadrant = list()
self.frequency = []
circle_list = SpecialCircleList()
d = 0
bound_up = 0
bound_down = 0
bound_left = 0
bound_right = 0
max_score = 0
def main():
global circle_list
global max_score
global d
global bound_up
global bound_down
global bound_left
global bound_right
line_in = input("")
(n, d) = line_in.split()
n = int(n)
d = int(d)
for i in range(n):
line_in = input("")
(x, y, r) = line_in.split()
x = int(x)
y = int(y)
r = int(r)
newCircle = Circle(x, y, r)
circle_list.append(newCircle)
#determine checking bounds
if(y+r > bound_up):
bound_up = y + r
if(y-r < bound_down):
bound_down = y - r
if(x + r > bound_right):
bound_right = x + r
if(x - r < bound_left):
bound_left = x - r
#brute force for angles 0->360
for angle in range(0, 360, 1):
#generate unit vector with direction
unit_vector = Point(math.cos(angle*math.pi/180),math.sin(angle*math.pi/180))
score_for_line = checkIfInAnyCircles(unit_vector)
if(score_for_line > max_score):
max_score = score_for_line
print(max_score)
#returns the score of a line/trajectory
def checkIfInAnyCircles(unit_vector):
global circle_list
global d
list_of_relavent_circles, max_d = circle_list.getRelevantList(unit_vector)
score = 0
if(len(list_of_relavent_circles) < (max_score/2)):
return(0)
bounce_vector = Point(unit_vector.x * d, unit_vector.y * d)
b = Point(bounce_vector.x, bounce_vector.y)
max_d = int((max_d + 50)/d)
i = 0
while(i < max_d):
for c in list_of_relavent_circles:
if(b.x >= c.A.x and b.y >= c.A.y and b.x <= c.B.x and b.y <= c.B.y):
if(thoroughCheck(b, c)):
score += 1
b.x += bounce_vector.x
b.y += bounce_vector.y
i += 1
return(score)
#after passing a basic bounded box collision check, do a more thorough check with
#floating point arithmatic to see if a bounce (b) lands within a given circle, (c)
def thoroughCheck(b, c):
distanceFromCircleCentre = (b.x - c.x)**2 + (b.y - c.y)**2
distanceFromCircleCentre = math.sqrt(distanceFromCircleCentre)
if(distanceFromCircleCentre <= c.r):
return(True)
else:
return(False)
if __name__ == "__main__":
main()
``` | instruction | 0 | 6,171 | 3 | 12,342 |
No | output | 1 | 6,171 | 3 | 12,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
n = int(input())
p = 10000
for i in range(n):
x, d = input().split()
x = int(x)
if d == 'South':
if p != -10000 and p - x >= -10000:
p -= x
else:
print('NO')
exit()
elif d == 'North':
if p != 10000 and p + x <= 10000:
p += x
else:
print('NO')
exit()
else:
if p == 10000 or p == -10000:
print('NO')
exit()
if p == 10000:
print('YES')
else:
print('NO')
``` | instruction | 0 | 6,265 | 3 | 12,530 |
Yes | output | 1 | 6,265 | 3 | 12,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
n = int(input())
l = []
for i in range(n):
a , b = map(str, input().split())
a = int(a)
c = [a, b]
l.append(c)
milles = 0
for i in range(n):
dist = l[i][0]
direcc = l[i][1]
if direcc == "South":
milles = milles + dist
elif direcc == "North":
milles = milles - dist
elif (direcc == "West" or direcc == "East") and (milles == 0 or milles == 20000):
print("NO")
exit(0)
if milles < 0 or milles > 2 * 10**4:
print("NO")
exit(0)
if milles != 0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 6,266 | 3 | 12,532 |
Yes | output | 1 | 6,266 | 3 | 12,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
import sys
n=int(input())
y=0
for i in range(n):
t,d=input().split()
t=int(t)
if y<0 or y>20000:
print('NO')
exit()
if d[0]=='N':
y-=t
elif d[0]=='S':
y+=t
elif y==0 or y==20000:
print('NO')
exit()
print('YES' if y==0 else 'NO')
``` | instruction | 0 | 6,267 | 3 | 12,534 |
Yes | output | 1 | 6,267 | 3 | 12,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
n = int(input())
def solve(n):
pos = 0
for i in range(n):
km, direc = input().split()
km = int(km)
if pos == 0 and direc != "South":
print("NO")
return
elif pos == 20000 and direc != "North":
print("NO")
return
elif direc == "South":
pos += km
if pos > 20000:
print("NO")
return
elif direc == "North":
pos -= km
if pos < 0:
print("NO")
return
if pos == 0:
print("YES")
else:
print("NO")
solve(n)
``` | instruction | 0 | 6,268 | 3 | 12,536 |
Yes | output | 1 | 6,268 | 3 | 12,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
total = 0
for _ in range(int(input().strip())):
dist, direction = input().strip().split()
dist = int(dist)
if dist%20000 == 0 and dist != 20000:
dist = 0
else:
dist = 20000-dist%20000 if dist > 20000 else dist
if total == 0:
if direction[0] != 'S':
print('NO')
quit()
if total == 20000:
if direction[0] != 'N':
print('NO')
quit()
if direction[0] == 'S':
total += dist
if direction[0] == 'N':
total -= dist
print(['YES', 'NO'][total!=0])
``` | instruction | 0 | 6,269 | 3 | 12,538 |
No | output | 1 | 6,269 | 3 | 12,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
cury = 0
for i in range(n):
k, d = input().split()
k = int(k)
if cury == 0 and d != "South":
print("NO")
exit()
if cury == 20000 and d != "North":
print("NO")
exit()
if d == "East" or d == "West":
continue
if d == "North":
k = k%40000
cury -= k
if cury < 0:
cury += 40000
cury = cury % 40000
if cury >= 20000:
cury = 40000 - cury
if d == "South":
k = k%40000
cury += k
cury = cury % 40000
if cury >= 20000:
cury = 40000 - cury
if cury%40000 != 0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 6,270 | 3 | 12,540 |
No | output | 1 | 6,270 | 3 | 12,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
n=int(input().strip())
det=20000
for _ in range(n):
dis,dir=input().strip().split()
dis=int(dis)%40000
if 20000<dis:
dis-=40000
if det==20000 and dir!='North':
det-=dis
elif det==0 and dir!='South':
det+=dis
elif dir=='South' and det!=0:
det-=dis
elif dir=='North' and det!=20000:
det+=dis
if det>20000:
det=40000-det
if det<0:
det=-det
print('YES' if det==20000 else 'NO')
``` | instruction | 0 | 6,271 | 3 | 12,542 |
No | output | 1 | 6,271 | 3 | 12,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem we assume the Earth to be a completely round ball and its surface a perfect sphere. The length of the equator and any meridian is considered to be exactly 40 000 kilometers. Thus, travelling from North Pole to South Pole or vice versa takes exactly 20 000 kilometers.
Limak, a polar bear, lives on the North Pole. Close to the New Year, he helps somebody with delivering packages all around the world. Instead of coordinates of places to visit, Limak got a description how he should move, assuming that he starts from the North Pole. The description consists of n parts. In the i-th part of his journey, Limak should move ti kilometers in the direction represented by a string diri that is one of: "North", "South", "West", "East".
Limak isn’t sure whether the description is valid. You must help him to check the following conditions:
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the North Pole, he can move only to the South.
* If at any moment of time (before any of the instructions or while performing one of them) Limak is on the South Pole, he can move only to the North.
* The journey must end on the North Pole.
Check if the above conditions are satisfied and print "YES" or "NO" on a single line.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 50).
The i-th of next n lines contains an integer ti and a string diri (1 ≤ ti ≤ 106, <image>) — the length and the direction of the i-th part of the journey, according to the description Limak got.
Output
Print "YES" if the description satisfies the three conditions, otherwise print "NO", both without the quotes.
Examples
Input
5
7500 South
10000 East
3500 North
4444 West
4000 North
Output
YES
Input
2
15000 South
4000 East
Output
NO
Input
5
20000 South
1000 North
1000000 West
9000 North
10000 North
Output
YES
Input
3
20000 South
10 East
20000 North
Output
NO
Input
2
1000 North
1000 South
Output
NO
Input
4
50 South
50 North
15000 South
15000 North
Output
YES
Note
Drawings below show how Limak's journey would look like in first two samples. In the second sample the answer is "NO" because he doesn't end on the North Pole.
<image>
Submitted Solution:
```
Q=int(input())
X=0
for _ in range(Q):
A=input().strip().split(' ')
D=int(A[0])
if X==0:
if A[1][0]!='S':
print('NO')
break
else:
X+=D
X%=40000
if X>20000:
X=40000-X
elif X==20000:
if A[1][0]!='N':
print('NO')
break
else:
X-=D
X%=40000
if X>20000:
X=40000-X
else:
if A[1][0]=='S':
X+=D
X%=40000
if X>20000:
X=40000-X
elif A[1][0]=='N':
X-=D
X%=40000
if X>20000:
X=40000-X
else:
if X%40000==0:
print('YES')
else:
print('NO')
``` | instruction | 0 | 6,272 | 3 | 12,544 |
No | output | 1 | 6,272 | 3 | 12,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,317 | 3 | 12,634 |
Tags: geometry
Correct Solution:
```
R,x1,y1,x2,y2=list(map(float,input().split()))
d=((x2-x1)**2+(y2-y1)**2)**0.5
if d>R:
print(x1," ",y1," ",R)
elif d==0:
print(x1-R/2," ",y1," ",R/2)
else:
x,y=(x1*(R+d)-x2*R)/(d),(y1*(R+d)-y2*R)/(d)
xans,yans=(x+x2)/2,(y+y2)/2
r=((x-xans)**2+(y-yans)**2)**0.5
print(xans," ",yans," ",r)
``` | output | 1 | 6,317 | 3 | 12,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,318 | 3 | 12,636 |
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2=map(int,input().split())
if ((x1-x2)**2+(y1-y2)**2>r*r):
print(x1,y1,r)
exit()
d=((x1-x2)**2+(y1-y2)**2)**0.5
r2=(r+d)/2
if (d==0):
x3=x2+r/2
y3=y2
else:
x3=x2+(x1-x2)/d*r2
y3=y2+(y1-y2)/d*r2
print(x3,y3,r2)
``` | output | 1 | 6,318 | 3 | 12,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,319 | 3 | 12,638 |
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2=map(int,input().split())
if (x2-x1)**2+(y2-y1)**2>=r*r:
print(x1,y1,r)
exit()
if x1==x2:
if y2<y1:
print(x1,y2+(y1+r-y2)/2,(y1+r-y2)/2)
else:
print(x1,y2-(y2-y1+r)/2,(y2-y1+r)/2)
else:
k=(y2-y1)/(x2-x1)
b=y1-k*x1
B=b+0
a=k*k+1
c=x1*x1+b*b-2*b*y1+y1*y1-r*r
b=2*k*b-2*x1-2*k*y1
d=b*b-4*a*c
X1=(-b+d**0.5)/(2*a)
X2=(-b-d**0.5)/(2*a)
Y1=k*X1+B
Y2=k*X2+B
r1=((X1-x2)**2+(Y1-y2)**2)**0.5
r2=((X2-x2)**2+(Y2-y2)**2)**0.5
if r1>r2:
print((x2+X1)/2,(y2+Y1)/2,r1/2)
else:
print((x2+X2)/2,(y2+Y2)/2,r2/2)
``` | output | 1 | 6,319 | 3 | 12,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,320 | 3 | 12,640 |
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2=map(int ,input().split())
re = ((x1-x2)**2+(y1-y2)**2)**0.5
if (re>=r):
re = r
x3 = x1
y3 = y1
if (re==0):
x3 = x2 + ((re+r)/2);
y3 = y2 ;
re = ((re+r)/2);
else:
x3 = x2 + ((((re+r)/2)/re) * (x1-x2));
y3 = y2 + ((((re+r)/2)/re) * (y1-y2));
re = ((re+r)/2);
print(x3,y3,re)
``` | output | 1 | 6,320 | 3 | 12,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,321 | 3 | 12,642 |
Tags: geometry
Correct Solution:
```
r,x1,y1,x2,y2 = map(int,input().split())
ttt = 0
import math
if(pow(x2-x1,2)+pow(y2-y1,2) >= r*r):
rad = r;
print(x1,' ',y1,' ',rad)
else:
rad=(r + math.sqrt(pow(x2-x1,2)+pow(y2-y1,2)))/2
dist= (math.sqrt(pow(x2-x1,2)+pow(y2-y1,2)))
m = rad - dist
if dist!=0:
x = ((m+dist)*x1 -(m)*x2)/dist
y = ((m+dist)*y1 -(m)*y2)/dist
else:
x = x1+r/2
y = y1
rad = 0.5*r
print(x,' ',y,' ',rad)
``` | output | 1 | 6,321 | 3 | 12,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,322 | 3 | 12,644 |
Tags: geometry
Correct Solution:
```
import math
par = input()
# par = '10 0 0 0 0'
def dist(x1, y1, x2, y2):
return math.sqrt((x1-x2)** 2 + (y1-y2)** 2)
def setap(R, x1, y1, x2, y2):
d = dist(x1, y1, x2, y2)
if d < R and d > 0:
r = ( d + R ) / 2
x = x2 - r / d * (x2 - x1)
y = y2 - r / d * (y2 - y1)
elif d == 0:
x = x2 + R / 2
y = y2
r = R / 2
else:
r = R
x = x1
y = y1
return [x,y,r]
par = list(map(int, par.split()))
R = par[0]
x1 = par[1]
y1 = par[2]
x2 = par[3]
y2 = par[4]
res = setap(R, x1, y1, x2, y2)
print(res[0],res[1], res[2])
``` | output | 1 | 6,322 | 3 | 12,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,323 | 3 | 12,646 |
Tags: geometry
Correct Solution:
```
# Reads two numbers from input and typecasts them to int using
# list comprehension
import math
R, x0, y0, x1, y1 = [int(x) for x in input().split()]
d1=math.sqrt((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0))
f=0
if((x1-x0)*(x1-x0)+(y1-y0)*(y1-y0)>R*R):
f=1
r=(d1+R)/2;
if(x1!=x0):
ten=(y1-y0)/(x1-x0)
if(ten>0):
if(y1>y0):
sn=math.sin(math.atan(ten))
ansy=y1-(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1-(cn*r)
else:
sn=math.sin(math.atan(ten))
ansy=y1+(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1+(cn*r)
else:
ten=abs(ten)
if(y1>y0):
sn=math.sin(math.atan(ten))
ansy=y1-(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1+(cn*r)
else:
sn=math.sin(math.atan(ten))
ansy=y1+(sn*r)
cn=math.cos(math.atan(ten))
ansx=x1-(cn*r)
else:
ansx=x1
if(y1>y0):
ansy=y1-r
else:
ansy=y1+r
if(f==0):
print(ansx, ansy, r)
else:
print(x0, y0, R)
``` | output | 1 | 6,323 | 3 | 12,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0 | instruction | 0 | 6,324 | 3 | 12,648 |
Tags: geometry
Correct Solution:
```
R,x1,y1,x2,y2=map(int,input().split(' '))
d1=(x2-x1)**2+(y2-y1)**2
if d1>=R*R:
print(str(x1)+" "+str(y1)+" "+str(R))
else:
d2=R+d1**(0.5)
if x1==x2 and y1==y2:
print(str(x1+R/2)+" "+str(y1)+" "+str(d2/2))
elif x1==x2:
if y2>y1:
print(str(x1)+" "+str((y1+y2-R)/2)+" "+str(d2/2))
else:
print(str(x1)+" "+str((y1+y2+R)/2)+" "+str(d2/2))
elif y1==y2:
if x2>x1:
print(str((x1+x2-R)/2)+" "+str(y1)+" "+str(d2/2))
else:
print(str((x1+x2+R)/2)+" "+str(y1)+" "+str(d2/2))
else:
x2-=x1
y2-=y1
slp=y2/x2
a=(R*R/(slp*slp+1))**(0.5)
b=slp*a
if x2<0 and y2<0:
print(str((a+x2)/2+x1)+" "+str((b+y2)/2+y1)+" "+str(d2/2))
elif x2<0 and y2>0:
print(str((a+x2)/2+x1)+" "+str((y2+b)/2+y1)+" "+str(d2/2))
elif x2>0 and y2<0:
print(str((x2-a)/2+x1)+" "+str((y2-b)/2+y1)+" "+str(d2/2))
else:
print(str((x2-a)/2+x1)+" "+str((y2-b)/2+y1)+" "+str(d2/2))
``` | output | 1 | 6,324 | 3 | 12,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
R, a, b, c, d = map(int, input().split())
da = ((a-c)**2 + (b-d)**2)**0.5
if da >= R:
print(a, b, R)
elif da == 0:
print(a, b+R/2, R/2)
else:
x, y = a - R*(c-a)/da, b-R*(d-b)/da
print((x+c)/2, (y+d)/2, (da+R)/2)
``` | instruction | 0 | 6,325 | 3 | 12,650 |
Yes | output | 1 | 6,325 | 3 | 12,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
import math
def main():
r1, x1, y1, x2, y2 = (int(x) for x in input().split())
if (x2 - x1) ** 2 + (y2 - y1) ** 2 >= r1 ** 2:
print(x1, y1, r1)
return
v1 = (x1 - x2, y1 - y2)
len_v1 = math.sqrt(v1[0] ** 2 + v1[1] ** 2)
r = (r1 + len_v1) / 2
if (x1, y1) == (x2, y2):
print(v1[0] + x2, v1[1] + r + y2, r)
return
k = (r1 + len_v1) / 2 / len_v1
v_res = (v1[0] * k + x2, v1[1] * k + y2)
print(v_res[0], v_res[1], r)
main()
``` | instruction | 0 | 6,326 | 3 | 12,652 |
Yes | output | 1 | 6,326 | 3 | 12,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
R, x1, y1, x2, y2 = map(int, input().split())
dx = x2 - x1
dy = y2 - y1
dlen = (dx**2 + dy**2)**0.5
if dlen < R:
if dlen == 0.0:
ax = x1 - R
ay = y1
else:
ax = x1 - dx * R/dlen
ay = y1 - dy * R/dlen
x = (ax + x2) / 2
y = (ay + y2) / 2
r = (R + dlen) / 2
else:
x = x1
y = y1
r = R
print("{:.15f} {:.15f} {:.15f}".format(x, y, r))
``` | instruction | 0 | 6,327 | 3 | 12,654 |
Yes | output | 1 | 6,327 | 3 | 12,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def vector_by_points(pt1, pt2):
return Vector(pt2.x - pt1.x, pt2.y - pt1.y)
class Vector:
def __init__(self, x, y):
self.x = x
self.y = y
def length(self):
return (self.x ** 2 + self.y ** 2) ** 0.5
def __add__(self, other):
return Vector(self.x + other.x, self.y + other.y)
def __mul__(self, other):
return Vector(self.x * other, self.y * other)
def normalise(self):
l = self.length()
if l != 0:
return Vector(self.x / l, self.y / l)
return self
r, x1, y1, x2, y2 = map(int, input().split())
center = Point(x1, y1)
notebook = Point(x2, y2)
v = vector_by_points(notebook, center)
dst = v.length()
if dst - r <= 0.000001:
ans = (2 * r - (r - dst)) / 2
if dst == 0:
print(x1 + r / 2, y1, ans)
else:
v = v.normalise() * ans + notebook
print(v.x, v.y, ans)
else:
print(x1, y1, r)
``` | instruction | 0 | 6,328 | 3 | 12,656 |
Yes | output | 1 | 6,328 | 3 | 12,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
R,x1,y1,x2,y2=input().split()
R=float(R)
x1=float(x1)
y1=float(y1)
x2=float(x2)
y2=float(y2)
d=((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))**0.5
k1=d;
k2=(R-d)/2;
if d==0:
print(x1,y1,0)
elif d<=R:
print((x1*(k1+k2)-k2*x2)/k1,(y1*(k1+k2)-k2*y2)/k1,(R+d)/2)
else:
print(x1,y1,R)
``` | instruction | 0 | 6,329 | 3 | 12,658 |
No | output | 1 | 6,329 | 3 | 12,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
import math
def dist(x1,y1,x2,y2):
return math.sqrt(distsqr(x1,y1,x2,y2))
def distsqr(x1,y1,x2,y2):
return ((x1-x2)**2 + (y1-y2)**2)
def f(x):
return math.sqrt(resr**2 - (x-x2)**2) + y2
"""
def bs(l, r):
t = 0
while(abs(l-r) > eps and t < 100):
mid = (l+r)/2
if(abs(dist(mid, f(mid), x2, y2)-resr) > eps):
l = mid
else:
r = mid
return (l+r)/2
"""
if __name__ == '__main__':
r, x1, y1, x2, y2 = map(int, input().split())
# print(r, x1, y1, x2, y2)
eps = 1e-16
d = distsqr(x1, y1, x2, y2)
if(d >= r*r):
print(x1, y1, r)
else:
resr = (math.sqrt(d)+r)/2
if y2 > y1:
yr = y2 - resr
else:
yr = y2 + resr
if x2 > x1:
xr = x2 - resr
else:
xr = x2 + resr
print(xr, yr, resr)
#print(dist(xr,yr,x1,y1), r)
``` | instruction | 0 | 6,330 | 3 | 12,660 |
No | output | 1 | 6,330 | 3 | 12,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
from math import sqrt
R, x1, y1, x2, y2 = map(int, input().split())
if (x2-x1)**2 + (y2-y1)**2 <= R*R:
mnR = sqrt((x2-x1)**2 + (y2-y1)**2)
r = (mnR + R)/2
if x1 == x2 and y1 == y2:
x, y = float(x2+((x1+R)-x1)/2), float(y2+((y1+R)-y1)/2)
else:
x = (x1-x2)*r/mnR + x2
y = (y1-y2)*r/mnR + y2
print(x, y, r)
else:
print("%.1f %.1f %.1f" % (x1, y1, R))
``` | instruction | 0 | 6,331 | 3 | 12,662 |
No | output | 1 | 6,331 | 3 | 12,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.
The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.
Input
The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).
Output
Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range.
Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.
Examples
Input
5 3 3 1 1
Output
3.7677669529663684 3.7677669529663684 3.914213562373095
Input
10 5 5 5 15
Output
5.0 5.0 10.0
Submitted Solution:
```
import math
from math import sin,cos
from decimal import Decimal
R,x1,y1,x2,y2=map(int,input().split())
if(((x2-x1)**2+(y2-y1)**2)**0.5>=R):
print(x1,y1,R)
exit(0)
if(x2==x1):
theta=22/14
else:
theta=math.atan(abs((y2-y1)/(x2-x1)))
r=(((x2-x1)**2+(y2-y1)**2)**0.5+R)/2
if(x1==x2 and y1==y2):
print(x1,y1,"0")
exit(0)
if(x2<=x1 and y2<=y1):
x=(x1+R*cos(theta)+x2)/2.0
y=(y1+R*sin(theta)+y2)/2.0
elif(x2>=x1 and y2>=y1):
x=(x1-R*cos(theta)+x2)/2.0
y=(y1-R*sin(theta)+y2)/2.0
elif(x2<=x1 and y2>=y1):
x=(x1+R*cos(theta)+x2)/2.0
y=(y1-R*sin(theta)+y2)/2.0
elif(x2>=x1 and y2<=y1):
x=(x1-R*cos(theta)+x2)/2.0
y=(y1+R*sin(theta)+y2)/2.0
print(Decimal(x),Decimal(y),Decimal(r))
``` | instruction | 0 | 6,332 | 3 | 12,664 |
No | output | 1 | 6,332 | 3 | 12,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
from sys import stdin,stdout
from math import gcd,sqrt,factorial,pi
from collections import deque,defaultdict
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
lcm=lambda x,y:(x*y)//gcd(x,y)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:1 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
sm=lambda x:(x**2+x)//2
N=10**9+7
n,k=R()
mn=2**((k*8)//n)
k=0
lst=-1
a=[]
for i in sorted(R()):
if i!=lst:
a+=0,
k+=1
a[-1]+=1
lst=i
if k<=mn:print(0);exit()
ans=0
for i in range(mn):
ans+=a[i]
val=ans
for i in range(mn,k):
val+=a[i]
val-=a[i-mn]
ans=max(ans,val)
print(sum(a)-ans)
``` | instruction | 0 | 6,653 | 3 | 13,306 |
Yes | output | 1 | 6,653 | 3 | 13,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
def mp():
return map(int, input().split())
n, I = mp()
a = sorted(list(mp()))
k = 2 ** ((I * 8) // n)
cnt = [[a[0], 1]]
for i in range(1, n):
if cnt[-1][0] == a[i]:
cnt[-1][1] += 1
else:
cnt.append([a[i], 1])
N = n
n = len(cnt)
a = [i[1] for i in cnt]
s = sum(a[:k])
ans = s
for i in range(k, n):
s -= a[i - k]
s += a[i]
ans = max(ans, s)
print(sum(a) - ans)
``` | instruction | 0 | 6,654 | 3 | 13,308 |
Yes | output | 1 | 6,654 | 3 | 13,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
from collections import defaultdict as dd
n,I=map(int,input().split())
l=list(map(int,input().split()))
ll=[]
last=-1
l.sort()
for i in l:
if(i!=last):
ll.append(1)
last=i
else:
ll[-1]+=1
k=len(ll)
while k > (1 << ((8*I)//n)):
k -= 1
ans=0
for i in range(k):
ans+=ll[i]
res=ans
for i in range(k,len(ll)):
ans=ans+ll[i]-ll[i-k]
res=max(res,ans)
print(sum(ll)-res)
``` | instruction | 0 | 6,655 | 3 | 13,310 |
Yes | output | 1 | 6,655 | 3 | 13,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
import math as mt
import collections as cc
import sys
I=lambda:list(map(int,input().split()))
n,i=I()
l=I()
l.sort()
tot=2**((8*i)//n)
ans=-10
temp=[0]
for i in range(1,n):
if l[i]!=l[i-1]:
temp.append(i)
if len(temp)<=tot:
print(0)
else:
x=-1
for i in range(len(temp)-tot):
x=max(temp[i+tot]-temp[i],x)
print(n-x)
``` | instruction | 0 | 6,656 | 3 | 13,312 |
Yes | output | 1 | 6,656 | 3 | 13,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
import math
n,I=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
I*=8
k=1<<(I//n)
dp=[1]
for i in range(1,n):
if a[i]==a[i-1]:
dp[-1]+=1
else:
dp.append(1)
if k>=len(dp):
print(0)
else:
ans=sum(dp[:k])
for i in range(k,len(dp)):
ans=max(ans,ans+dp[i]-dp[i-k])
print(n-ans)
``` | instruction | 0 | 6,657 | 3 | 13,314 |
No | output | 1 | 6,657 | 3 | 13,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
n,k=map(int,input().split())
l=list(map(int,input().split()))
se=set(l)
r=pow(2,(8*k)//n)
d=dict()
for i in range(len(l)):
if l[i] not in d:
d.update({l[i]:1})
else:
d[l[i]]+=1
freq=[]
for i in sorted(d):
freq.append(d[i])
s=0
e=len(freq)-1
k=len(se)
for i in range(1,len(freq)):
freq[i]+=freq[i-1]
ans=999999999999999999
#print(freq)
freq=[0]+freq
for i in range(k-r+1):
rt=freq[i]+freq[-1]-freq[len(freq)-1+i-k+r]
print(rt,freq[i])
ans=min(ans,rt)
print(ans)
``` | instruction | 0 | 6,658 | 3 | 13,316 |
No | output | 1 | 6,658 | 3 | 13,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
#TO MAKE THE PROGRAM FAST
''' ---------------------------------------------------------------------------------------------------- '''
import sys
import heapq
import math
input = sys.stdin.readline
sys.setrecursionlimit(100000)
from collections import *
''' ---------------------------------------------------------------------------------------------------- '''
#FOR TAKING INPUTS
''' ---------------------------------------------------------------------------------------------------- '''
def li():return [int(____) for ____ in input().rstrip('\n').split(' ')]
def val():return int(input().rstrip('\n'))
def st():return input().rstrip('\n')
def sttoli():return [int(i) for i in input().rstrip('\n')]
''' ---------------------------------------------------------------------------------------------------- '''
#MAIN PROGRAM
n,I=li()
l=li()
I*=8
cnt = Counter()
d = deque()
l.sort()
# print(l)
currlen = currdistinct = 0
i, ans = 0, -1
while i<len(l):
cnt[l[i]] += 1
currlen += 1
if not len(d) or d[-1] != l[i]:
d.append(l[i])
currdistinct += 1
while len(d) and n * math.ceil(math.log2(currdistinct)) > I:
d.popleft()
currlen -= cnt[d[0]]
currdistinct -= 1
ans = max(ans,currlen)
i+=1
print(len(l)-ans)
''' ---------------------------------------------------------------------------------------------------- '''
``` | instruction | 0 | 6,659 | 3 | 13,318 |
No | output | 1 | 6,659 | 3 | 13,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One common way of digitalizing sound is to record sound intensity at particular time moments. For each time moment intensity is recorded as a non-negative integer. Thus we can represent a sound file as an array of n non-negative integers.
If there are exactly K distinct values in the array, then we need k = ⌈ log_{2} K ⌉ bits to store each value. It then takes nk bits to store the whole file.
To reduce the memory consumption we need to apply some compression. One common way is to reduce the number of possible intensity values. We choose two integers l ≤ r, and after that all intensity values are changed in the following way: if the intensity value is within the range [l;r], we don't change it. If it is less than l, we change it to l; if it is greater than r, we change it to r. You can see that we lose some low and some high intensities.
Your task is to apply this compression in such a way that the file fits onto a disk of size I bytes, and the number of changed elements in the array is minimal possible.
We remind you that 1 byte contains 8 bits.
k = ⌈ log_{2} K ⌉ is the smallest integer such that K ≤ 2^{k}. In particular, if K = 1, then k = 0.
Input
The first line contains two integers n and I (1 ≤ n ≤ 4 ⋅ 10^{5}, 1 ≤ I ≤ 10^{8}) — the length of the array and the size of the disk in bytes, respectively.
The next line contains n integers a_{i} (0 ≤ a_{i} ≤ 10^{9}) — the array denoting the sound file.
Output
Print a single integer — the minimal possible number of changed elements.
Examples
Input
6 1
2 1 2 3 4 3
Output
2
Input
6 2
2 1 2 3 4 3
Output
0
Input
6 1
1 1 2 2 3 3
Output
2
Note
In the first example we can choose l=2, r=3. The array becomes 2 2 2 3 3 3, the number of distinct elements is K=2, and the sound file fits onto the disk. Only two values are changed.
In the second example the disk is larger, so the initial file fits it and no changes are required.
In the third example we have to change both 1s or both 3s.
Submitted Solution:
```
n, I = map(int, input().split())
a = sorted([x for x in map(int, input().split())])
a=[-1]+a
b=[i for i in range(n) if a[i] < a[i+1]]
ans = list()
for x in zip(b, b[2**I:]+[n]):
ans.append(x[1]-x[0])
print(n-max(ans))
``` | instruction | 0 | 6,660 | 3 | 13,320 |
No | output | 1 | 6,660 | 3 | 13,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible. | instruction | 0 | 6,695 | 3 | 13,390 |
Tags: binary search, brute force, math
Correct Solution:
```
t1,t2,x1,x2,t0=map(int,input().split())
y1,y2,a1,a2,m=x1,x2,0,0,10000000
while True:
t=(t1*y1+t2*y2)/(y1+y2)
if t>=t0:
if t-t0<m:a1,a2,m=y1,y2,t-t0
if y1==0 or y2==0 or t==t0:break
if t>t0:
y2-=1
else:
y1-=1
print(a1,a2)
``` | output | 1 | 6,695 | 3 | 13,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible. | instruction | 0 | 6,696 | 3 | 13,392 |
Tags: binary search, brute force, math
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 998244353
def factors(n):return sorted(list(set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def compare(a,b):
if a[-1]<b[-1]:return -1
elif a[-1]>b[-1]:return 1
else:
if sum(a[:-1]) > sum(b[:-1]):return -1
return 1
def give(a,b):
temp = (a*t1 + b*t2)/(a + b)
# print(temp,a,b,t)
if temp < t:
return [a,b,float('inf')]
return [a,b,abs(t - temp)]
t1, t2, x1, x2, t = li()
bestans = []
if t == t1 and t1 == t2:
print(x1,x2)
elif t == t1:
print(x1,0)
elif t == t2:
print(0,x2)
else:
currans = []
currcheck = float('inf')
r = (t2 - t)/(t - t1)
# print('R ; ',r)
bestans.append(give(x1,x2))
bestans.append(give(0,x2))
bestans.append(give(x1,0))
for i in bestans:
if currcheck > i[-1]:
currcheck = i[-1]
currans = i[:-1]
elif currcheck == i[-1] and sum(currans) < sum(i[:-1]):
currans = i[:-1]
do = 1
for i in range(1,x2 + 1):
curry1 = i*r
a = curry1
b = i
if 0 <= int(curry1) <= x1:
do = 0
c,d,e = give(int(a),b)
if e < currcheck:
currcheck = e
currans = [c,d]
elif e == currcheck and sum(currans) < c + d:
currans = [c,d]
if 0 <= int(a) + 1 <= x1:
do = 0
c,d,e = give(int(a) + 1,b)
if e < currcheck:
currcheck = e
currans = [c,d]
elif e == currcheck and sum(currans) < c + d:
currans = [c,d]
if do:
for j in range(1,min(x1,x2) + 1):
c,d,e = give(j,j)
if e < currcheck:
currcheck = e
currans = [c,d]
elif e == currcheck and sum(currans) < c + d:
currans = [c,d]
# for i,(a,b,c) in enumerate(bestans):
# if ((a*t1 + b*t2)/(a + b)) < t:
# bestans[i][-1] = 1000
# continue
# bestans[i][-1] = abs(t - ((a*t1 + b*t2)/(a + b)))
# bestans.sort(key=cmp_to_key(compare))
# print(bestans)
print(*currans)
``` | output | 1 | 6,696 | 3 | 13,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible. | instruction | 0 | 6,697 | 3 | 13,394 |
Tags: binary search, brute force, math
Correct Solution:
```
def gcd(a,b):
if(a%b==0):
return b
else:
return gcd(b,a%b)
[t1,t2,x1,x2,t0]=input().split(' ')
#[t1,t2,x1,x2,t0]=[100,110,2,2,109]
t1=int(t1)
t2=int(t2)
x1=int(x1)
x2=int(x2)
t0=int(t0)
a=t2-t0
b=t0-t1
if(a==0 and b==0):
print(x1,x2)
exit(0)
if (a==0):
y1=0
y2=x2
print(y1,y2)
exit(0)
if (b==0):
y1=x1
y2=0
print(y1,y2)
exit(0)
g=gcd(a,b)
a=int(a/g)
b=int(b/g)
if(a<=x1 and b<=x2):
mintime=int(x1/a)
if(mintime>int(x2/b)):
mintime=int(x2/b)
print(mintime*a,mintime*b)
exit(0)
y1=1
y2=1
miny1=1
miny2=1
minn=99999
fg=0
while(y1<=x1 and y2<=x2):
if(y1/y2<a/b):
if(minn>a/b-y1/y2):
minn=a/b-y1/y2
miny1=y1
miny2=y2
fg=1
y1=y1+1
else:
y2=y2+1
y1=miny1
y2=miny2
mintime=int(x1/y1)
if(mintime>int(x2/y2)):
mintime=int(x2/y2)
if(fg==1):
print(mintime*y1,mintime*y2)
else:
print(0,x2)
``` | output | 1 | 6,697 | 3 | 13,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible. | instruction | 0 | 6,698 | 3 | 13,396 |
Tags: binary search, brute force, math
Correct Solution:
```
import math
def gcd(a,b):
if(b==0):
return a
return gcd(b,a%b)
l=input().split()
t1=int(l[0])
t2=int(l[1])
x1=int(l[2])
x2=int(l[3])
t0=int(l[4])
num1=t2-t0
num2=t0-t1
if(t1==t2):
print(x1,x2)
quit()
if(num1==0):
print(0,x2)
quit()
if(num2==0):
print(x1,0)
quit()
z=num2/num1
maxa=10**18
ans=(0,0)
for i in range(1,x1+1):
ok=z*i
if(ok>x2):
break
num1=i
num2=math.ceil(ok)
if(maxa==((num2/num1)-z) and num2+num1>ans[0]+ans[1]):
ans=(num1,num2)
elif(maxa>((num2/num1)-z)):
ans=(num1,num2)
maxa=((num2/num1-z))
if(ans==(0,0)):
ans=(0,x2)
print(ans[0],ans[1])
``` | output | 1 | 6,698 | 3 | 13,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible. | instruction | 0 | 6,699 | 3 | 13,398 |
Tags: binary search, brute force, math
Correct Solution:
```
import math
t1,t2,x1,x2,t0=(int(i) for i in input().split())
t=(0,x2)
m=x2
t3=t2
if(t0==t2):
if(t1==t2):
print(x1,x2)
else:
print(t[0],t[1])
else:
if([t1,t2,x1,x2,t0]==[1,1000000,1000000,1000000,999999]):
print(1,999998)
else:
for i in range(1,x1+1):
k=(t1-t0)*i/(t0-t2)
k1=(t1*i+t2*math.ceil(k))/(i+math.ceil(k))
if(k1<=t2 and k<=x2):
if(k1==t3 and k1>=t0):
if(i+math.ceil(k)>m):
m=i+math.ceil(k)
t=(i,math.ceil(k))
elif(k1<t3 and k1>=t0):
t3=k1
m=i+math.ceil(k)
t=(i,math.ceil(k))
k2=(t1*i+t2*int(k))/(i+int(k))
if(k2==t3 and k2>=t0):
if(i+int(k)>m):
m=i+int(k)
t=(i,int(k))
elif(k2<t3 and k2>=t0):
t3=k2
m=i+int(k)
t=(i,int(k))
print(t[0],t[1])
``` | output | 1 | 6,699 | 3 | 13,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 998244353
def factors(n):return sorted(list(set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def compare(a,b):
if a[-1]<b[-1]:return -1
elif a[-1]>b[-1]:return 1
else:
if sum(a[:-1]) > sum(b[:-1]):return -1
return 1
def give(a,b):
temp = (a*t1 + b*t2)/(a + b)
if temp < t:
return [a,b,1000]
return [a,b,abs(t - temp)]
t1, t2, x1, x2, t = li()
bestans = []
if t == t1 and t1 == t2:
print(x1,x2)
elif t == t1:
print(x1,0)
elif t == t2:
print(0,x2)
else:
currans = []
currcheck = float('inf')
r = (t2 - t)/(t - t1)
bestans.append(give(x1,x2))
bestans.append(give(0,x2))
bestans.append(give(x1,0))
for i in bestans:
if currcheck > i[-1]:
currcheck = i[-1]
currans = i[:-1]
if currcheck == i[-1] and sum(currans) < sum(i[:-1]):
currans = i[:-1]
for i in range(1,x2 + 1):
curry1 = i*r
a = curry1
b = i
if 0 <= int(curry1) <= x1:
c,d,e = give(int(a),b)
if e < currcheck:
currcheck = e
currans = [c,d]
if e == currcheck and sum(currans) < c + d:
currans = [c,d]
if 0 <= math.ceil(curry1) <= x1:
c,d,e = give(math.ceil(a),b)
if e < currcheck:
currcheck = e
currans = [c,d]
if e == currcheck and sum(currans) < c + d:
currans = [c,d]
# for i,(a,b,c) in enumerate(bestans):
# if ((a*t1 + b*t2)/(a + b)) < t:
# bestans[i][-1] = 1000
# continue
# bestans[i][-1] = abs(t - ((a*t1 + b*t2)/(a + b)))
# bestans.sort(key=cmp_to_key(compare))
# print(bestans)
print(*currans)
``` | instruction | 0 | 6,700 | 3 | 13,400 |
No | output | 1 | 6,700 | 3 | 13,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible.
Submitted Solution:
```
a=input()[:2]
if (a=="10"):print("99 33")
else:print("0 1000")
``` | instruction | 0 | 6,701 | 3 | 13,402 |
No | output | 1 | 6,701 | 3 | 13,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible.
Submitted Solution:
```
def R(): return map(int, input().split())
def I(): return int(input())
def S(): return str(input())
def L(): return list(R())
from collections import Counter
import math
import sys
t1,t2,x1,x2,t0=R()
mint=10*100
y1min=-1
y2min=-1
for y1 in range(x1+1):
y2=min(math.ceil(y1*(t0-t1)/(t2-t0)),x2)
if y1+y2==0: continue
if mint>(y1*t1+y2*t2)/(y1+y2):
mint=(y1*t1+y2*t2)/(y1+y2)
#print(mint)
y1min=y1
y2min=y2
if y1min==0:
k=x2//y2min
elif y2min==0:
k=x1//y1min
else:
k=min(x1//y1min,x2//y2min)
print(k*y1min,k*y2min)
``` | instruction | 0 | 6,702 | 3 | 13,404 |
No | output | 1 | 6,702 | 3 | 13,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bob is about to take a hot bath.
There are two taps to fill the bath: a hot water tap and a cold water tap. The cold water's temperature is t1, and the hot water's temperature is t2. The cold water tap can transmit any integer number of water units per second from 0 to x1, inclusive. Similarly, the hot water tap can transmit from 0 to x2 water units per second.
If y1 water units per second flow through the first tap and y2 water units per second flow through the second tap, then the resulting bath water temperature will be:
<image>
Bob wants to open both taps so that the bath water temperature was not less than t0. However, the temperature should be as close as possible to this value. If there are several optimal variants, Bob chooses the one that lets fill the bath in the quickest way possible.
Determine how much each tap should be opened so that Bob was pleased with the result in the end.
Input
You are given five integers t1, t2, x1, x2 and t0 (1 ≤ t1 ≤ t0 ≤ t2 ≤ 106, 1 ≤ x1, x2 ≤ 106).
Output
Print two space-separated integers y1 and y2 (0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2).
Examples
Input
10 70 100 100 25
Output
99 33
Input
300 500 1000 1000 300
Output
1000 0
Input
143 456 110 117 273
Output
76 54
Note
In the second sample the hot water tap shouldn't be opened, but the cold water tap should be opened at full capacity in order to fill the bath in the quickest way possible.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 998244353
def factors(n):return sorted(list(set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def compare(a,b):
if a[-1]<b[-1]:return -1
elif a[-1]>b[-1]:return 1
else:
if sum(a[:-1]) > sum(b[:-1]):return -1
return 1
t1, t2, x1, x2, t = li()
bestans = []
if t == t1 and t1 == t2:
print(x1,x2)
elif t == t1:
print(x1,0)
elif t == t2:
print(0,x2)
else:
r = (t2 - t)/(t - t1)
# bestans.append([0,0,0])
bestans.append([0,x2,0])
bestans.append([x1,0,0])
for i in range(1,x2 + 1):
curry1 = i*r
if 0 <= int(curry1) <= x1:
bestans.append([int(curry1),i,curry1 - int(curry1)])
if 0 <= math.ceil(curry1) <= x1:
bestans.append([math.ceil(curry1),i,math.ceil(curry1) - curry1])
for i,(a,b,c) in enumerate(bestans):
if ((a*t1 + b*t2)/(a + b)) < t:
bestans[i][-1] = 1000
continue
bestans[i][-1] = abs(t - ((a*t1 + b*t2)/(a + b)))
bestans.sort(key=cmp_to_key(compare))
# print(bestans)
print(*bestans[0][:-1])
``` | instruction | 0 | 6,703 | 3 | 13,406 |
No | output | 1 | 6,703 | 3 | 13,407 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: every minute at each junction (x, y) containing at least four ants a group of four ants will be formed, and these four ants will scatter to the neighbouring junctions (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) — one ant in each direction. No other ant movements will happen. Ants never interfere with each other.
Scientists have put a colony of n ants into the junction (0, 0) and now they wish to know how many ants will there be at some given junctions, when the movement of the ants stops.
Input
First input line contains integers n (0 ≤ n ≤ 30000) and t (1 ≤ t ≤ 50000), where n is the number of ants in the colony and t is the number of queries. Each of the next t lines contains coordinates of a query junction: integers xi, yi ( - 109 ≤ xi, yi ≤ 109). Queries may coincide.
It is guaranteed that there will be a certain moment of time when no possible movements can happen (in other words, the process will eventually end).
Output
Print t integers, one per line — the number of ants at the corresponding junctions when the movement of the ants stops.
Examples
Input
1 3
0 1
0 0
0 -1
Output
0
1
0
Input
6 5
0 -2
0 -1
0 0
0 1
0 2
Output
0
1
2
1
0
Note
In the first sample the colony consists of the one ant, so nothing happens at all.
In the second sample the colony consists of 6 ants. At the first minute 4 ants scatter from (0, 0) to the neighbouring junctions. After that the process stops. | instruction | 0 | 6,950 | 3 | 13,900 |
Tags: brute force, implementation
Correct Solution:
```
from sys import *
f = lambda: map(int, stdin.readline().split())
n, t = f()
m = 65
r = range(m)
p = [[0] * m for i in r]
p[1][0] = n // 4
p[0][0] = n % 4
q = k = 1
while q:
k += 1
q = 0
for x in r[1:k]:
for y in r[:x + 1]:
if p[x][y] < 4: continue
q = 1
d = p[x][y] // 4
p[x][y] %= 4
p[x + 1][y] += d
if x > y:
if x > y + 1:
p[x][y + 1] += d
p[x - 1][y] += d
else:
p[x][x] += 2 * d
if y: p[y][y] += 2 * d
else: p[x][y] += d
if y: p[x][y - 1] += d if y > 1 else 2 * d
s = []
for j in range(t):
x, y = f()
x, y = abs(x), abs(y)
if x < y: x, y = y, x
s.append(p[x][y] if x < m else 0)
stdout.write('\n'.join(map(str, s)))
``` | output | 1 | 6,950 | 3 | 13,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: every minute at each junction (x, y) containing at least four ants a group of four ants will be formed, and these four ants will scatter to the neighbouring junctions (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) — one ant in each direction. No other ant movements will happen. Ants never interfere with each other.
Scientists have put a colony of n ants into the junction (0, 0) and now they wish to know how many ants will there be at some given junctions, when the movement of the ants stops.
Input
First input line contains integers n (0 ≤ n ≤ 30000) and t (1 ≤ t ≤ 50000), where n is the number of ants in the colony and t is the number of queries. Each of the next t lines contains coordinates of a query junction: integers xi, yi ( - 109 ≤ xi, yi ≤ 109). Queries may coincide.
It is guaranteed that there will be a certain moment of time when no possible movements can happen (in other words, the process will eventually end).
Output
Print t integers, one per line — the number of ants at the corresponding junctions when the movement of the ants stops.
Examples
Input
1 3
0 1
0 0
0 -1
Output
0
1
0
Input
6 5
0 -2
0 -1
0 0
0 1
0 2
Output
0
1
2
1
0
Note
In the first sample the colony consists of the one ant, so nothing happens at all.
In the second sample the colony consists of 6 ants. At the first minute 4 ants scatter from (0, 0) to the neighbouring junctions. After that the process stops. | instruction | 0 | 6,951 | 3 | 13,902 |
Tags: brute force, implementation
Correct Solution:
```
from sys import *
f = lambda: map(int, stdin.readline().split())
n, t = f()
m = 65
r = range(1, m)
p = [[0] * m for i in range(m)]
p[1][0] = n // 4
p[0][0] = n % 4
s = k = 1
while s:
s = 0
for x in r[:k]:
if p[x][0] > 3:
s = 1
d = p[x][0] // 4
p[x][0] %= 4
p[x + 1][0] += d
p[x][1] += d
if x != 1:
p[x - 1][0] += d
else:
p[1][1] += d
p[1][0] += d
for y in r[:x - 1]:
if p[x][y] > 3:
s = 1
d = p[x][y] // 4
p[x][y] %= 4
p[x + 1][y] += d
p[x - 1][y] += d
p[x][y + 1] += d
p[x][y - 1] += d
if x == y + 1:
p[x][x] += d
p[y][y] += d
if y == 1: p[x][0] += d
if p[x][x] > 3:
s = 1
d = p[x][x] // 4
p[x][x] %= 4
p[x + 1][x] += d
p[x][x - 1] += d
if x == 1: p[1][0] += d
k += 1
s = []
for j in range(t):
x, y = f()
x, y = abs(x), abs(y)
if x < y: x, y = y, x
s.append(p[x][y] if x < m else 0)
stdout.write('\n'.join(map(str, s)))
``` | output | 1 | 6,951 | 3 | 13,903 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: every minute at each junction (x, y) containing at least four ants a group of four ants will be formed, and these four ants will scatter to the neighbouring junctions (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) — one ant in each direction. No other ant movements will happen. Ants never interfere with each other.
Scientists have put a colony of n ants into the junction (0, 0) and now they wish to know how many ants will there be at some given junctions, when the movement of the ants stops.
Input
First input line contains integers n (0 ≤ n ≤ 30000) and t (1 ≤ t ≤ 50000), where n is the number of ants in the colony and t is the number of queries. Each of the next t lines contains coordinates of a query junction: integers xi, yi ( - 109 ≤ xi, yi ≤ 109). Queries may coincide.
It is guaranteed that there will be a certain moment of time when no possible movements can happen (in other words, the process will eventually end).
Output
Print t integers, one per line — the number of ants at the corresponding junctions when the movement of the ants stops.
Examples
Input
1 3
0 1
0 0
0 -1
Output
0
1
0
Input
6 5
0 -2
0 -1
0 0
0 1
0 2
Output
0
1
2
1
0
Note
In the first sample the colony consists of the one ant, so nothing happens at all.
In the second sample the colony consists of 6 ants. At the first minute 4 ants scatter from (0, 0) to the neighbouring junctions. After that the process stops. | instruction | 0 | 6,952 | 3 | 13,904 |
Tags: brute force, implementation
Correct Solution:
```
from sys import *
f = lambda: map(int, stdin.readline().split())
n, t = f()
m = 65
r = range(m)
p = [[0] * m for i in r]
p[1][0] = n // 4
p[0][0] = n % 4
ultima = 5
q = k = 1
while q:
k += 1
q = 0
for x in r[1:k]:
for y in r[:x + 1]:
if p[x][y] < 4: continue
q = 1
d = p[x][y] // 4
p[x][y] %= 4
p[x + 1][y] += d
if x > y:
if x > y + 1:
p[x][y + 1] += d
p[x - 1][y] += d
else:
p[x][x] += 2 * d
if y: p[y][y] += 2 * d
else: p[x][y] += d
if y: p[x][y - 1] += d if y > 1 else 2 * d
s = []
for j in range(t):
x, y = f()
x, y = abs(x), abs(y)
if x < y: x, y = y, x
s.append(p[x][y] if x < m else 0)
stdout.write('\n'.join(map(str, s)))
``` | output | 1 | 6,952 | 3 | 13,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: every minute at each junction (x, y) containing at least four ants a group of four ants will be formed, and these four ants will scatter to the neighbouring junctions (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) — one ant in each direction. No other ant movements will happen. Ants never interfere with each other.
Scientists have put a colony of n ants into the junction (0, 0) and now they wish to know how many ants will there be at some given junctions, when the movement of the ants stops.
Input
First input line contains integers n (0 ≤ n ≤ 30000) and t (1 ≤ t ≤ 50000), where n is the number of ants in the colony and t is the number of queries. Each of the next t lines contains coordinates of a query junction: integers xi, yi ( - 109 ≤ xi, yi ≤ 109). Queries may coincide.
It is guaranteed that there will be a certain moment of time when no possible movements can happen (in other words, the process will eventually end).
Output
Print t integers, one per line — the number of ants at the corresponding junctions when the movement of the ants stops.
Examples
Input
1 3
0 1
0 0
0 -1
Output
0
1
0
Input
6 5
0 -2
0 -1
0 0
0 1
0 2
Output
0
1
2
1
0
Note
In the first sample the colony consists of the one ant, so nothing happens at all.
In the second sample the colony consists of 6 ants. At the first minute 4 ants scatter from (0, 0) to the neighbouring junctions. After that the process stops.
Submitted Solution:
```
from collections import deque
n, t = map(int,input().split()) # at (0,0) there are n ants
g = [[0]*201 for x in range(201)]
base = 100
g[base][base] = n
x = deque()
y = deque()
while len(x) > 0:
curX = x.popleft()
curY = y.popleft()
if g[curX][curY] < 4:
continue
re = g[curX][curY] // 4
g[curX][curY] %= 4
g[curX][curY+1] += re
x.append(curX)
y.append(curY+1)
g[curX][curY-1] += re
x.append(curX)
y.append(curY-1)
g[curX+1][curY] += re
x.append(curX+1)
y.append(curY)
g[curX-1][curY] += re
x.append(curX-1)
y.append(curY)
x.append(curX)
y.append(curY)
# def broadcast(x,y):
# if g[x][y] >= 4:
# re = g[x][y] // 4
# g[x][y] = g[x][y] % 4
# g[x-1][y] += re
# g[x+1][y] += re
# g[x][y+1] += re
# g[x][y-1] += re
# else:
# return
# broadcast(x+1,y);
# broadcast(x-1,y);
# broadcast(x,y+1);
# broadcast(x,y-1);
# broadcast(x,y);
# broadcast(base,base)
while t:
x, y = map(int,input().split())
if x > 10 or x < -10 or y > 10 or y < -10:
print(0)
else:
print(g[x+base][y+base])
t -= 1
``` | instruction | 0 | 6,953 | 3 | 13,906 |
No | output | 1 | 6,953 | 3 | 13,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: every minute at each junction (x, y) containing at least four ants a group of four ants will be formed, and these four ants will scatter to the neighbouring junctions (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) — one ant in each direction. No other ant movements will happen. Ants never interfere with each other.
Scientists have put a colony of n ants into the junction (0, 0) and now they wish to know how many ants will there be at some given junctions, when the movement of the ants stops.
Input
First input line contains integers n (0 ≤ n ≤ 30000) and t (1 ≤ t ≤ 50000), where n is the number of ants in the colony and t is the number of queries. Each of the next t lines contains coordinates of a query junction: integers xi, yi ( - 109 ≤ xi, yi ≤ 109). Queries may coincide.
It is guaranteed that there will be a certain moment of time when no possible movements can happen (in other words, the process will eventually end).
Output
Print t integers, one per line — the number of ants at the corresponding junctions when the movement of the ants stops.
Examples
Input
1 3
0 1
0 0
0 -1
Output
0
1
0
Input
6 5
0 -2
0 -1
0 0
0 1
0 2
Output
0
1
2
1
0
Note
In the first sample the colony consists of the one ant, so nothing happens at all.
In the second sample the colony consists of 6 ants. At the first minute 4 ants scatter from (0, 0) to the neighbouring junctions. After that the process stops.
Submitted Solution:
```
from copy import deepcopy
a = [[0] * 30 for _ in range(30)]
b = deepcopy(a)
n, t = map(int, input().split(' '))
a[0][0] = n
for i in range(20):
for j in range(-15, 16):
for k in range(-15, 16):
if (a[j][k] >= 4):
ea = a[j][k] // 4
left = a[j][k] - 4 * ea
b[j][k] += left
b[j][k+1] += ea
b[j][k-1] += ea
b[j-1][k] += ea
b[j+1][k] += ea
else:
b[j][k] += a[j][k]
a = deepcopy(b)
b = [[0] * 30 for _ in range(30)]
for i in range(t):
x, y = map(int, input().split(' '))
if abs(x) >= 20 or abs(y) >= 20:
print(0)
else:
print(a[x][y])
``` | instruction | 0 | 6,954 | 3 | 13,908 |
No | output | 1 | 6,954 | 3 | 13,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: every minute at each junction (x, y) containing at least four ants a group of four ants will be formed, and these four ants will scatter to the neighbouring junctions (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) — one ant in each direction. No other ant movements will happen. Ants never interfere with each other.
Scientists have put a colony of n ants into the junction (0, 0) and now they wish to know how many ants will there be at some given junctions, when the movement of the ants stops.
Input
First input line contains integers n (0 ≤ n ≤ 30000) and t (1 ≤ t ≤ 50000), where n is the number of ants in the colony and t is the number of queries. Each of the next t lines contains coordinates of a query junction: integers xi, yi ( - 109 ≤ xi, yi ≤ 109). Queries may coincide.
It is guaranteed that there will be a certain moment of time when no possible movements can happen (in other words, the process will eventually end).
Output
Print t integers, one per line — the number of ants at the corresponding junctions when the movement of the ants stops.
Examples
Input
1 3
0 1
0 0
0 -1
Output
0
1
0
Input
6 5
0 -2
0 -1
0 0
0 1
0 2
Output
0
1
2
1
0
Note
In the first sample the colony consists of the one ant, so nothing happens at all.
In the second sample the colony consists of 6 ants. At the first minute 4 ants scatter from (0, 0) to the neighbouring junctions. After that the process stops.
Submitted Solution:
```
import sys
import time
_no_of_ants, _no_of_queries = (input().split())
no_of_ants = int(_no_of_ants)
no_of_queries = int(_no_of_queries)
class ants_cord():
def __init__(self,point,no_of_ants):
self.co_ant = {point:no_of_ants}
def ants_at_point(self,point):
return self.co_ant[point]
def move_ants(self,point):
x1,y1 = point
if (x1+1,y1) in self.co_ant.keys():
self.co_ant[((x1+1),y1)] += 1
else:
self.co_ant[((x1+1),y1)] = 1
if (x1,y1+1) in self.co_ant.keys():
self.co_ant[((x1),y1+1)] += 1
else:
self.co_ant[((x1),y1+1)] = 1
if (x1,y1-1) in self.co_ant.keys():
self.co_ant[((x1),y1-1)] += 1
else:
self.co_ant[((x1),y1-1)] = 1
if (x1-1,y1) in self.co_ant.keys():
self.co_ant[((x1-1),y1)] += 1
else:
self.co_ant[((x1-1),y1)] = 1
self.co_ant[point] -= 4
# NEED TO OPTIMIZE THIS SECTION
# CAUSING TLE
def check_status(self):
for key in self.co_ant.keys():
if k.co_ant[key] >= 4:
return key
else: return False
start = time.time()
k = ants_cord((0,0),no_of_ants)
point = k.check_status()
while point:
k.move_ants(point)
point = k.check_status()
end = time.time()
print("Sec:{}".format(end-start))
while no_of_queries:
_x,_y = input().split()
x = int(_x)
y = int(_y)
if (x,y) in k.co_ant:
print(k.co_ant[(x,y)])
else:
print('0')
no_of_queries -= 1
``` | instruction | 0 | 6,955 | 3 | 13,910 |
No | output | 1 | 6,955 | 3 | 13,911 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.