message
stringlengths 2
23.4k
| message_type
stringclasses 2
values | message_id
int64 0
1
| conversation_id
int64 129
108k
| cluster
float64 6
6
| __index_level_0__
int64 258
216k
|
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,010
| 6
| 206,020
|
Tags: dp
Correct Solution:
```
m=1000000007
s=input()
if(s.count('w') or s.count('m')):
print(0)
else:
dp=[1,1]
n=len(s)
for i in range(1,n):
if(s[i]==s[i-1] and (s[i]=='u' or s[i]=='n')):
dp.append((dp[i]+dp[i-1])%m)
else:
dp.append(dp[i]%m)
#print(dp)
print(dp[-1]%m)
```
|
output
| 1
| 103,010
| 6
| 206,021
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,011
| 6
| 206,022
|
Tags: dp
Correct Solution:
```
f = [1, 2]
const = 10 ** 9 + 1
s = input()
if "m" in s or "w" in s:
print(0)
else:
ans = 1
const = 10 ** 9 + 7
check = False
count = 0
if s[0] == "u" or s[0] == "n":
check = True
count = 1
for i in range(1, len(s)):
f.append((f[-1] + f[-2]) % const)
if s[i] == s[i - 1]:
count += 1
elif check:
ans = (ans * f[count - 1]) % const
count = 0
check = False
if s[i] == "u":
count = 1
check = "u"
elif s[i] == "n":
count = 1
check = "n"
else:
check = False
count = 0
elif s[i] == "u":
count = 1
check = "u"
elif s[i] == "n":
count = 1
check = "n"
else:
check = False
count = 0
if check:
ans = (ans * f[count - 1]) % const
print(ans)
```
|
output
| 1
| 103,011
| 6
| 206,023
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,012
| 6
| 206,024
|
Tags: dp
Correct Solution:
```
# import sys
# input = sys.stdin.readline
s = input()
n = len(s)
A = []
mod = 10**9+7
L = [0,1,2]
for i in range(10**5+3):
L.append((L[-1]+L[-2]) % mod)
nn=0
uu=0
for i in range(n):
if s[i] == "n":
if nn == 0:
nn = 1
A.append(1)
else:
A[-1] += 1
else:
nn = 0
if s[i] == "u":
if uu == 0:
uu = 1
A.append(1)
else:
A[-1] += 1
else:
uu = 0
ans = 1
for i in range(len(A)):
ans = (ans * L[A[i]]) % mod
if s.count("m") > 0 or s.count("w") > 0:
ans = 0
print(ans)
```
|
output
| 1
| 103,012
| 6
| 206,025
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,013
| 6
| 206,026
|
Tags: dp
Correct Solution:
```
mod = 10**9+7
def cmb(n, r, mod=mod):
if ( r<0 or r>n ):
return 0
r = min(r, n-r)
return g1[n] * g2[r] * g2[n-r] % mod
NN = 10**5 # 使うデータによって変える
g1 = [1, 1] # 元テーブル
g2 = [1, 1] #逆元テーブル
inverse = [0, 1] #逆元テーブル計算用テーブル
for i in range( 2, NN + 1 ):
g1.append( ( g1[-1] * i ) % mod )
inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
g2.append( (g2[-1] * inverse[-1]) % mod )
import sys
input = sys.stdin.readline
S = input().rstrip()
N = len(S)
pre = '-1'
ans = 1
A = []
c = 0
for s in S:
if s == 'm' or s == 'w':
ans = 0
break
if s == 'n':
if pre == 'n':
c += 1
else:
if c > 1:
A.append(c)
c = 1
elif s == 'u':
if pre == 'u':
c += 1
else:
if c > 1:
A.append(c)
c = 1
else:
if c > 1:
A.append(c)
c = 0
pre = s
A.append(c)
if ans == 0:
print(0)
else:
for a in A:
tmp = 1
n, m = a-1, 1
while n >= m:
tmp += cmb(n, m)
tmp %= mod
n -= 1
m += 1
ans = ans * tmp % mod
print(ans)
```
|
output
| 1
| 103,013
| 6
| 206,027
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,014
| 6
| 206,028
|
Tags: dp
Correct Solution:
```
s = input()
s_ct = 0
n_ct = 0
ans = []
dp = [0 for _ in range(len(s)+5)]
dp[0] = 1
dp[1] = 1
dp[2] = 2
a = 1
for i in range(3,len(dp)):
dp[i] = (dp[i-1]+dp[i-2])%(10**9+7 )
for i in range(len(s)):
if(s[i]=='m'):
a = 0
if(s[i]=='w'):
a=0
if(s[i]=='u'):
if n_ct:
ans.append(n_ct)
n_ct = 0
s_ct+=1
elif s[i] == 'n':
if s_ct:
ans.append(s_ct)
s_ct = 0
n_ct+=1
else:
if s_ct:
ans.append(s_ct)
s_ct = 0
if n_ct:
ans.append(n_ct)
n_ct = 0
if s_ct:
ans.append(s_ct)
s_ct = 0
if n_ct:
ans.append(n_ct)
n_ct = 0
for i in ans:
a = (a*dp[i])%(10**9 + 7)
print(a)
```
|
output
| 1
| 103,014
| 6
| 206,029
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,015
| 6
| 206,030
|
Tags: dp
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
import bisect
import math
# import itertools
# import heapq
# from queue import PriorityQueue, LifoQueue, SimpleQueue
# import sys.stdout.flush() use for interactive problems
alpha = 'abcdefghijklmnopqrstuvwxyz'
ALPHA = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
inf = 1e17
mod = 10 ** 9 + 7
# Max = 10**6
# primes = []
# prime = [True for i in range(10**6+1)]
# p = 2
# while (p * p <= Max+1):
#
# # If prime[p] is not
# # changed, then it is a prime
# if (prime[p] == True):
#
# # Update all multiples of p
# for i in range(p * p, Max+1, p):
# prime[i] = False
# p += 1
#
# for p in range(2, Max+1):
# if prime[p]:
# primes.append(p)
def factorial(n):
f = 1
for i in range(1, n + 1):
f = (f * i) % mod # Now f never can
# exceed 10^9+7
return f
def ncr(n, r):
# initialize numerator
# and denominator
num = den = 1
for i in range(r):
num = (num * (n - i)) % mod
den = (den * (i + 1)) % mod
return (num * pow(den,
mod - 2, mod)) % mod
def solve(s):
if 'w' in s or 'm' in s:
return 0
dp = [1,1]
for i in range(1,len(s)) :
if s[i] == s[i-1] == 'u' or s[i] == s[i-1] == 'n':
dp.append((dp[-1]+dp[-2])%mod)
else:
dp.append(dp[-1])
return dp[-1] % mod
pass
t = 1#int(input())
ans = []
for _ in range(1):
# n = int(input())
# n,k = map(int, input().split())
# arr = [int(x) for x in input().split()]
# queries = [int(x) for x in input().split()]
# arr = list(input())
s = input()
# t = input()
# customers = []
# for i in range(n):
# customers.append([int(x) for x in input().split()])
# k = int(input())
# s = [int(x) for x in input().split()]
# qs = []
# for j in range(q):
# r,c = map(int,input().split())
# qs.append((r,c))
ans.append(solve(s))
for j in range(len(ans)):
#print('Case #' + str(j + 1) + ": " + str(ans[j]))
print(ans[j])
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
|
output
| 1
| 103,015
| 6
| 206,031
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,016
| 6
| 206,032
|
Tags: dp
Correct Solution:
```
s = input()
L = [0] * (len(s) + 1)
inf = 1000000007
L[0] = 1
L[1] = 1
for i in range(1, len(s)):
if s[i] == 'm' or s[i] == 'w':
break
if s[i] == 'u':
if s[i - 1] == 'u':
L[i + 1] = (L[i] + L[i - 1]) % inf
else:
L[i + 1] = L[i]
elif s[i] == 'n':
if s[i - 1] == 'n':
L[i + 1] = (L[i] + L[i - 1]) % inf
else:
L[i + 1] = L[i]
else:
L[i + 1] = L[i]
if s[0] == 'm' or s[0] == 'w':
L[len(s)] = 0
print(L[len(s)])
```
|
output
| 1
| 103,016
| 6
| 206,033
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Constanze is the smartest girl in her village but she has bad eyesight.
One day, she was able to invent an incredible machine! When you pronounce letters, the machine will inscribe them onto a piece of paper. For example, if you pronounce 'c', 'o', 'd', and 'e' in that order, then the machine will inscribe "code" onto the paper. Thanks to this machine, she can finally write messages without using her glasses.
However, her dumb friend Akko decided to play a prank on her. Akko tinkered with the machine so that if you pronounce 'w', it will inscribe "uu" instead of "w", and if you pronounce 'm', it will inscribe "nn" instead of "m"! Since Constanze had bad eyesight, she was not able to realize what Akko did.
The rest of the letters behave the same as before: if you pronounce any letter besides 'w' and 'm', the machine will just inscribe it onto a piece of paper.
The next day, I received a letter in my mailbox. I can't understand it so I think it's either just some gibberish from Akko, or Constanze made it using her machine. But since I know what Akko did, I can just list down all possible strings that Constanze's machine would have turned into the message I got and see if anything makes sense.
But I need to know how much paper I will need, and that's why I'm asking you for help. Tell me the number of strings that Constanze's machine would've turned into the message I got.
But since this number can be quite large, tell me instead its remainder when divided by 10^9+7.
If there are no strings that Constanze's machine would've turned into the message I got, then print 0.
Input
Input consists of a single line containing a string s (1 ≤ |s| ≤ 10^5) — the received message. s contains only lowercase Latin letters.
Output
Print a single integer — the number of strings that Constanze's machine would've turned into the message s, modulo 10^9+7.
Examples
Input
ouuokarinn
Output
4
Input
banana
Output
1
Input
nnn
Output
3
Input
amanda
Output
0
Note
For the first example, the candidate strings are the following: "ouuokarinn", "ouuokarim", "owokarim", and "owokarinn".
For the second example, there is only one: "banana".
For the third example, the candidate strings are the following: "nm", "mn" and "nnn".
For the last example, there are no candidate strings that the machine can turn into "amanda", since the machine won't inscribe 'm'.
|
instruction
| 0
| 103,017
| 6
| 206,034
|
Tags: dp
Correct Solution:
```
import sys
# def recur(pre,p,pos,maxx,temp):
# if pos == len(pre):
# maxx[0] = max(maxx[0],temp)
# return
# for i in range(0,p+1):
# if i < len(pre[pos]) :
# temp = temp + pre[pos][i]
# recur(pre,p-i,pos+1,maxx,temp)
# temp = temp - pre[pos][i]
RI = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
# for _ in range(int(ri())):
mod = 10**9+7
st = ri()
dp = [0]*2
dp[1] = 1
for i in range(len(st)):
if st[i] == 'm' or st[i] == 'w':
dp[1] = 0
break
if st[i] == 'n' and st[i-1] == 'n':
temp = dp[1]
dp[1] = (dp[0]+dp[1])%mod
dp[0] = temp
elif st[i] == 'u' and st[i-1] == 'u':
temp = dp[1]
dp[1] = (dp[0]+dp[1])%mod
dp[0] = temp
else:
dp[0] = dp[1]
print(dp[1])
```
|
output
| 1
| 103,017
| 6
| 206,035
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,966
| 6
| 207,932
|
Tags: combinatorics, dp
Correct Solution:
```
from functools import reduce
n = int(input())
s1, s2 = str(input()), str(input())
print((10 ** sum([(s1[i] == '?') + (s2[i] == '?') for i in range(n)]) - (not reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)], False)) * reduce(lambda x, y: (x * y) % 1000000007, [55 if s1[i] == '?' and s2[i] == '?' else (ord(s2[i]) - ord('0') + 1) if s1[i] == '?' else (10 - ord(s1[i]) + ord('0')) if s2[i] == '?' else 1 for i in range(n)], 1) - (not reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)], False)) * reduce(lambda x, y: (x * y) % 1000000007, [55 if s1[i] == '?' and s2[i] == '?' else (10 - ord(s2[i]) + ord('0')) if s1[i] == '?' else (ord(s1[i]) - ord('0')) + 1 if s2[i] == '?' else 1 for i in range(n)], 1) + (not reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)], False) and not reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)], False)) * reduce(lambda x, y: (x * y) % 1000000007, [10 if s1[i] == '?' and s2[i] == '?' else 1 for i in range(n)], 1)) % 1000000007)
```
|
output
| 1
| 103,966
| 6
| 207,933
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,967
| 6
| 207,934
|
Tags: combinatorics, dp
Correct Solution:
```
n = int(input())
C = 10**9+7
a = input()
s = input()
d = [[0, 0, 0, 1]]
for q in range(n):
d.append(d[-1][::])
if a[q] != '?' and s[q] != '?':
if int(a[q]) < int(s[q]):
d[-1][1] = d[-1][3] = 0
d[-1][0] += d[-2][3]
d[-1][2] += d[-2][1]
elif int(a[q]) > int(s[q]):
d[-1][0] = d[-1][3] = 0
d[-1][1] += d[-2][3]
d[-1][2] += d[-2][0]
elif s[q] != '?':
d[-1][2] *= 10
d[-1][0] += d[-2][3]
d[-1][1] += d[-2][3]
d[-1][0] *= int(s[q])+1
d[-1][1] *= 10-int(s[q])
d[-1][0] -= d[-2][3]
d[-1][1] -= d[-2][3]
d[-1][2] += d[-2][0]*(9-int(s[q]))+d[-2][1]*int(s[q])
elif a[q] != '?':
d[-1][2] *= 10
d[-1][0] += d[-2][3]
d[-1][1] += d[-2][3]
d[-1][1] *= int(a[q]) + 1
d[-1][0] *= 10 - int(a[q])
d[-1][0] -= d[-2][3]
d[-1][1] -= d[-2][3]
d[-1][2] += d[-2][1] * (9 - int(a[q])) + d[-2][0] * int(a[q])
else:
d[-1][3] *= 10
d[-1][2] *= 100
d[-1][0] *= 55
d[-1][1] *= 55
d[-1][0] += d[-2][3]*45
d[-1][1] += d[-2][3]*45
d[-1][2] += d[-2][0]*45+d[-2][1]*45
d[-1][0] %= C
d[-1][1] %= C
d[-1][2] %= C
d[-1][3] %= C
print(d[-1][2])
```
|
output
| 1
| 103,967
| 6
| 207,935
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,968
| 6
| 207,936
|
Tags: combinatorics, dp
Correct Solution:
```
from functools import reduce
n = int(input())
s1, s2 = str(input()), str(input())
b1 = reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)], False)
b2 = reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)], False)
s = sum([(s1[i] == '?') + (s2[i] == '?') for i in range(n)])
ans1 = reduce(lambda x, y: (x * y) % 1000000007, [55 if s1[i] == '?' and s2[i] == '?' else (ord(s2[i]) - ord('0') + 1) if s1[i] == '?' else (10 - ord(s1[i]) + ord('0')) if s2[i] == '?' else 1 for i in range(n)], 1)
ans2 = reduce(lambda x, y: (x * y) % 1000000007, [55 if s1[i] == '?' and s2[i] == '?' else (10 - ord(s2[i]) + ord('0')) if s1[i] == '?' else (ord(s1[i]) - ord('0')) + 1 if s2[i] == '?' else 1 for i in range(n)], 1)
ans3 = reduce(lambda x, y: (x * y) % 1000000007, [10 if s1[i] == '?' and s2[i] == '?' else 1 for i in range(n)], 1)
print((10 ** s - (not b2) * ans1 - (not b1) * ans2 + (not b1 and not b2) * ans3) % 1000000007)
# Made By Mostafa_Khaled
```
|
output
| 1
| 103,968
| 6
| 207,937
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,969
| 6
| 207,938
|
Tags: combinatorics, dp
Correct Solution:
```
#!/usr/bin/python3
def build(n, s, t):
ans = 1
for i in range(n):
if s[i] == '?' and t[i] == '?':
ans = (55 * ans) % (10 ** 9 + 7)
elif s[i] == '?':
ans = ((ord(t[i]) - ord('0') + 1) * ans) % (10 ** 9 + 7)
elif t[i] == '?':
ans = ((ord('9') - ord(s[i]) + 1) * ans) % (10 ** 9 + 7)
return ans
n = int(input())
s = input()
t = input()
sltt = True
tlts = True
qm = 0
cqm = 0
for i in range(n):
if t[i] == '?':
qm += 1
if s[i] == '?':
qm += 1
if t[i] == '?' and s[i] == '?':
cqm += 1
if t[i] == '?' or s[i] == '?':
continue
if ord(s[i]) < ord(t[i]):
tlts = False
if ord(t[i]) < ord(s[i]):
sltt = False
if not sltt and not tlts:
print(pow(10, qm, 10 ** 9 + 7))
elif sltt and tlts:
print((pow(10, qm, 10 ** 9 + 7) - build(n, s, t) - build(n, t, s) + pow(10, cqm, 10 ** 9 + 7)) % (10 ** 9 + 7))
elif sltt:
print((pow(10, qm, 10 ** 9 + 7) - build(n, s, t)) % (10 ** 9 + 7))
else:
print((pow(10, qm, 10 ** 9 + 7) - build(n, t, s)) % (10 ** 9 + 7))
```
|
output
| 1
| 103,969
| 6
| 207,939
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,970
| 6
| 207,940
|
Tags: combinatorics, dp
Correct Solution:
```
import sys
mod = 1000000007
n = int(input())
s1 = input()
s2 = input()
flag1 = 0
flag2 = 0
for i in range(n):
if s1[i] != "?" and s2[i] != "?" and s1[i] > s2[i]:
flag1 = 1
if s1[i] != "?" and s2[i] != "?" and s1[i] < s2[i]:
flag2 = 1
if flag1 and flag2:
ans = 1
for i in range(n):
if s1[i] == "?":
ans *= 10
ans %= mod
for i in range(n):
if s2[i] == "?":
ans *= 10
ans %= mod
print (ans % mod)
sys.exit(0)
if not(flag1) and not(flag2):
ans1 = 1
ans2 = 1
x = 0
al = 1
for i in range(n):
if s1[i] == "?" and s2[i] == "?":
x += 1
for i in range(n):
if s1[i] == "?":
al *= 10
al %= mod
for i in range(n):
if s2[i] == "?":
al *= 10
al %= mod
for i in range(n):
if s1[i] == "?" and s2[i] == "?":
ans1 *= 55
ans1 %= mod
elif s1[i] == "?":
ans1 *= 10 - int(s2[i])
ans1 %= mod
elif s2[i] == "?":
ans1 *= int(s1[i]) + 1
ans1 %= mod
s1, s2 = s2, s1
for i in range(n):
if s1[i] == "?" and s2[i] == "?":
ans2 *= 55
ans2 %= mod
elif s1[i] == "?":
ans2 *= 10 - int(s2[i])
ans2 %= mod
elif s2[i] == "?":
ans2 *= int(s1[i]) + 1
ans2 %= mod
y = 1
for i in range(x):
y *= 10
y %= mod
print ((al - (ans1 + ans2 - y)) % mod)
sys.exit(0)
if flag2:
s1, s2 = s2, s1
ans = 1
al = 1
for i in range(n):
if s1[i] == "?":
al *= 10
al %= mod
for i in range(n):
if s2[i] == "?":
al *= 10
al %= mod
for i in range(n):
if s1[i] == "?" and s2[i] == "?":
ans *= 55
ans %= mod
elif s1[i] == "?":
ans *= 10 - int(s2[i])
ans %= mod
elif s2[i] == "?":
ans *= int(s1[i]) + 1
ans %= mod
print ((al - ans) % mod)
```
|
output
| 1
| 103,970
| 6
| 207,941
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,971
| 6
| 207,942
|
Tags: combinatorics, dp
Correct Solution:
```
from functools import reduce
n = int(input())
s1, s2 = str(input()), str(input())
b1 = reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)], False)
b2 = reduce(lambda x, y: x or y, [s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)], False)
s = sum([(s1[i] == '?') + (s2[i] == '?') for i in range(n)])
ans1 = reduce(lambda x, y: (x * y) % 1000000007, [55 if s1[i] == '?' and s2[i] == '?' else (ord(s2[i]) - ord('0') + 1) if s1[i] == '?' else (10 - ord(s1[i]) + ord('0')) if s2[i] == '?' else 1 for i in range(n)], 1)
ans2 = reduce(lambda x, y: (x * y) % 1000000007, [55 if s1[i] == '?' and s2[i] == '?' else (10 - ord(s2[i]) + ord('0')) if s1[i] == '?' else (ord(s1[i]) - ord('0')) + 1 if s2[i] == '?' else 1 for i in range(n)], 1)
ans3 = reduce(lambda x, y: (x * y) % 1000000007, [10 if s1[i] == '?' and s2[i] == '?' else 1 for i in range(n)], 1)
print((10 ** s - (not b2) * ans1 - (not b1) * ans2 + (not b1 and not b2) * ans3) % 1000000007)
```
|
output
| 1
| 103,971
| 6
| 207,943
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,972
| 6
| 207,944
|
Tags: combinatorics, dp
Correct Solution:
```
from functools import reduce
n, s1, s2, f1, f2 = int(input()), str(input()), str(input()), lambda x: reduce((lambda a, b: (a * b) % 1000000007), x, 1), lambda x: reduce((lambda a, b: a or b), x, False)
print((10 ** sum([(s1[i] == '?') + (s2[i] == '?') for i in range(n)]) - (not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)])) * f1([55 if s1[i] == '?' and s2[i] == '?' else (ord(s2[i]) - ord('0') + 1) if s1[i] == '?' else (10 - ord(s1[i]) + ord('0')) if s2[i] == '?' else 1 for i in range(n)]) - (not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)])) * f1([55 if s1[i] == '?' and s2[i] == '?' else (10 - ord(s2[i]) + ord('0')) if s1[i] == '?' else (ord(s1[i]) - ord('0')) + 1 if s2[i] == '?' else 1 for i in range(n)]) + (not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) < ord(s2[i]) for i in range(n)]) and not f2([s1[i] != '?' and s2[i] != '?' and ord(s1[i]) > ord(s2[i]) for i in range(n)])) * f1([10 if s1[i] == '?' and s2[i] == '?' else 1 for i in range(n)])) % 1000000007)
```
|
output
| 1
| 103,972
| 6
| 207,945
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string that consists of digits and question marks ("?").
Yaroslav has two string templates, each of them has length n. Yaroslav wants to count the number of ways to replace all question marks by some integers in both templates, so as to make the resulting strings incomparable. Note that the obtained strings can contain leading zeroes and that distinct question marks can be replaced by distinct or the same integers.
Help Yaroslav, calculate the remainder after dividing the described number of ways by 1000000007 (109 + 7).
Input
The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format.
Output
In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7).
Examples
Input
2
90
09
Output
1
Input
2
11
55
Output
0
Input
5
?????
?????
Output
993531194
Note
The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0.
|
instruction
| 0
| 103,973
| 6
| 207,946
|
Tags: combinatorics, dp
Correct Solution:
```
n = int(input())
a = input()
b = input()
mod = int(1e9+7)
x, y, z = 1, 1, 1
for i in range(n):
if a[i] == '?' and b[i] == '?':
x = (x * 55) % mod
y = (y * 55) % mod
z = (z * 10) % mod
elif a[i] == '?':
x = (x * (10 - int(b[i]))) % mod
y = (y * (int(b[i]) + 1)) % mod
elif b[i] == '?':
x = (x * (int(a[i]) + 1)) % mod
y = (y * (10 - int(a[i]))) % mod
else:
if int(a[i]) < int(b[i]): x = 0
if int(a[i]) > int(b[i]): y = 0
if a[i] != b[i]: z = 0
res = pow(10, a.count('?') + b.count('?'), mod)
print((res - x - y + z) % mod)
```
|
output
| 1
| 103,973
| 6
| 207,947
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,704
| 6
| 211,408
|
Tags: implementation, strings
Correct Solution:
```
s = str(input())
ll = sum(map(str.islower, s))
ul = sum(map(str.isupper, s))
if ul > ll:
print(s.upper())
else:
print(s.lower())
```
|
output
| 1
| 105,704
| 6
| 211,409
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,705
| 6
| 211,410
|
Tags: implementation, strings
Correct Solution:
```
a = input()
l = len(a)
count = 0
for i in a:
if i.isupper():
count += 1
if count <= int(l/2):
print(a.lower())
else:
print(a.upper())
```
|
output
| 1
| 105,705
| 6
| 211,411
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,706
| 6
| 211,412
|
Tags: implementation, strings
Correct Solution:
```
s=list(input())
p=""
l=[]
x=[]
for i in range(len(s)):
if s[i]==s[i].upper():
l.append(s[i])
elif s[i]==s[i].lower():
x.append(s[i])
if len(l)>len(x):
print(p.join(s).upper())
else:
print(p.join(s).lower())
```
|
output
| 1
| 105,706
| 6
| 211,413
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,707
| 6
| 211,414
|
Tags: implementation, strings
Correct Solution:
```
word = input()
uppercases = 0
lowercases = 0
for letter in word:
if letter.upper() == letter:
uppercases += 1
else:
lowercases += 1
if uppercases > lowercases:
print(word.upper())
else:
print(word.lower())
```
|
output
| 1
| 105,707
| 6
| 211,415
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,708
| 6
| 211,416
|
Tags: implementation, strings
Correct Solution:
```
s=input();print(s.upper() if sum(1 for c in s if c.isupper())>sum(1 for c in s if c.islower())else s.lower())
```
|
output
| 1
| 105,708
| 6
| 211,417
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,709
| 6
| 211,418
|
Tags: implementation, strings
Correct Solution:
```
w = input()
upp = 0
low = 0
for i in w:
if i.isupper() == True:
upp += 1
else:
low += 1
if upp <= low:
print(w.lower())
else:
print(w.upper())
```
|
output
| 1
| 105,709
| 6
| 211,419
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,710
| 6
| 211,420
|
Tags: implementation, strings
Correct Solution:
```
n=input()
l=0
u=0
for i in range(len(n)):
if n[i].islower():
l+=1
elif n[i].isupper():
u+=1
if l>u:
print(n.lower())
elif l<u:
print(n.upper())
else:
print(n.lower())
```
|
output
| 1
| 105,710
| 6
| 211,421
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input
The first line contains a word s — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output
Print the corrected word s. If the given word s has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Examples
Input
HoUse
Output
house
Input
ViP
Output
VIP
Input
maTRIx
Output
matrix
|
instruction
| 0
| 105,711
| 6
| 211,422
|
Tags: implementation, strings
Correct Solution:
```
word = str(input())
sumUppercaseLetters = 0
sumLowercaseLetters = 0
for i in word:
if i.isupper():
sumUppercaseLetters += 1
elif i.islower():
sumLowercaseLetters += 1
if sumUppercaseLetters > sumLowercaseLetters:
print(word.upper())
elif sumLowercaseLetters > sumUppercaseLetters:
print(word.lower())
else:
print(word.lower())
```
|
output
| 1
| 105,711
| 6
| 211,423
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,508
| 6
| 213,016
|
Tags: expression parsing, implementation
Correct Solution:
```
s = input()
i = s[1:].find('at') + 1
s = s[:i] + '@' + s[i+2:]
while True:
i = s[1:].find('dot')
if i == -1:
break
i += 1
if i+3 == len(s):
break
s = s[:i] + '.' + s[i+3:]
print(s)
```
|
output
| 1
| 106,508
| 6
| 213,017
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,509
| 6
| 213,018
|
Tags: expression parsing, implementation
Correct Solution:
```
s=input()
at=s.find('at',1)
s=s[:at]+'@'+s[at+2:]
pos=1
while 1:
ns=''
dot=s.find('dot',pos,-1)
if dot==-1:
break
ns+=s[:dot]
ns+='.'
pos=dot+1
ns+=s[dot+3:]
s=ns
print(s)
```
|
output
| 1
| 106,509
| 6
| 213,019
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,510
| 6
| 213,020
|
Tags: expression parsing, implementation
Correct Solution:
```
s=input()
if s[:3]=="dot" or s[:2]=="at" or s[len(s)-2:]=="at" or s[len(s)-3:]=="dot":
if s[:3]=="dot":
t=s[3:]
a=t.split('dot')
l=".".join(a)
b=l.find("at")
o=l[:b]+"@"+l[b+2:]
q="dot"+o
if q[-1]==".":
a=q[:len(q)-1]
s=a+"dot"
elif q[-1]=="@":
a=q[:len(q)-1]
s=a+"at"
else:
s=q
if s[:2]=="at":
t=s[2:]
a=t.split('dot')
l=".".join(a)
b=l.find('at')
o=l[:b]+"@"+l[b+2:]
q="at"+o
if q[-1]==".":
a=q[:len(q)-1]
s=a+"dot"
elif q[-1]=="@":
a=q[:len(q)-1]
s=a+"at"
else:
s=q
if s[len(s)-3:]=="dot" and s[:3]!="dot" and s[:2]!="at":
t=s[:len(s)-3]
a=t.split('dot')
l=".".join(a)
b=l.find('at')
q=l[:b]+"@"+l[b+2:]
s=q+"dot"
if s[len(s)-2:]=="at" and s[:3]!="dot" and s[:2]!="at":
t=s[:len(s)-2]
a=t.split('dot')
l=".".join(a)
b=l.find('at')
q=l[:b]+"@"+l[b+2:]
s=q+"at"
print(s)
else:
a=s.split('dot')
l=".".join(a)
b=l.find('at')
q=l[:b]+"@"+l[b+2:]
print(q)
```
|
output
| 1
| 106,510
| 6
| 213,021
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,511
| 6
| 213,022
|
Tags: expression parsing, implementation
Correct Solution:
```
from sys import stdin
input=stdin.readline
voiceEmail = input()
dotReplaced = voiceEmail[1:len(voiceEmail) - 2].replace('dot', '.')
voiceEmail = voiceEmail[0] + dotReplaced + voiceEmail[len(voiceEmail) - 2:]
atReplaced = voiceEmail[1:len(voiceEmail) - 2].replace('at', '@', 1)
voiceEmail = voiceEmail[0] + atReplaced + voiceEmail[len(voiceEmail) - 2:]
print(voiceEmail)
```
|
output
| 1
| 106,511
| 6
| 213,023
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,512
| 6
| 213,024
|
Tags: expression parsing, implementation
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
s = input().rstrip()
ans = s[0]
suf = s[-1]
s = s[1:-1]
at_flag = 0
while s:
if s[:3] == 'dot':
ans += '.'
s = s[3:]
elif not at_flag and s[:2] == 'at':
ans += '@'
at_flag = 1
s = s[2:]
else:
ans += s[0]
s = s[1:]
print(ans + suf)
```
|
output
| 1
| 106,512
| 6
| 213,025
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,513
| 6
| 213,026
|
Tags: expression parsing, implementation
Correct Solution:
```
import re
import sys
exit=sys.exit
from bisect import bisect_left as bsl,bisect_right as bsr
from collections import Counter,defaultdict as ddict,deque
from functools import lru_cache
cache=lru_cache(None)
from heapq import *
from itertools import *
from math import inf
from pprint import pprint as pp
enum=enumerate
ri=lambda:int(rln())
ris=lambda:list(map(int,rfs()))
rln=sys.stdin.readline
rl=lambda:rln().rstrip('\n')
rfs=lambda:rln().split()
cat=''.join
catn='\n'.join
mod=1000000007
d4=[(0,-1),(1,0),(0,1),(-1,0)]
d8=[(-1,-1),(0,-1),(1,-1),(-1,0),(1,0),(-1,1),(0,1),(1,1)]
########################################################################
def solve():
s=rl()
t=s[1:-1]
t=t.replace('at','@',1)
t=t.replace('dot','.')
return s[0]+t+s[-1]
print(solve())
```
|
output
| 1
| 106,513
| 6
| 213,027
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,514
| 6
| 213,028
|
Tags: expression parsing, implementation
Correct Solution:
```
def main():
desc = input()
chunk = desc[1:-1]
chunk = chunk.replace("at", "@", 1)
chunk = chunk.replace("dot", ".")
print(desc[0] + chunk + desc[-1])
main()
```
|
output
| 1
| 106,514
| 6
| 213,029
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
|
instruction
| 0
| 106,515
| 6
| 213,030
|
Tags: expression parsing, implementation
Correct Solution:
```
#!/bin/python
email = input()
m = {'@' : 'at', '.' : 'dot'}
replaced = ''
addAtLast = False
sep = []
def removeFirstLast(ch):
global replaced, addAtLast,sep, email
if m[ch] in email:
if sep[0] == '':
del sep[0]
replaced = m[ch] + sep[0]
else:
replaced = sep[0]
if len(sep)>0:
if sep[-1] == '':
del sep[-1]
addAtLast = True
else:
replaced = sep[0]
def addLast(ch) :
global replaced, addAtLast
if addAtLast :
replaced += m[ch]
#Code Starts from here :
#First replacing all the dots
sep = email.split(m['.'])
removeFirstLast('.')
for word in sep [1:]:
replaced += '.' + word
addLast('.')
#Next replacing all the ats
email = replaced
replaced = ''
addAtLast = False
sep = email.split(m['@'])
removeFirstLast('@')
#If there are more than 1 @ :
count = 0
for word in sep[1:] :
if count<1 :
replaced+= '@' + word
count+=1
else:
replaced+= 'at' + word
addLast('@')
print (replaced)
```
|
output
| 1
| 106,515
| 6
| 213,031
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
Submitted Solution:
```
s = input()
# don't start at first character!
# replace first occurance of 'at' with '@'
s = s[0] + s[1:].replace("at", "@", 1)
# replace the last 'dot' with '.'
s = s[0] + s[1:-1].replace("dot", ".") + s[-1]
print(s)
```
|
instruction
| 0
| 106,516
| 6
| 213,032
|
Yes
|
output
| 1
| 106,516
| 6
| 213,033
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
Submitted Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
def isInt(s):
return '0' <= s[0] <= '9'
MOD = 998244353
"""
"""
def solve():
S = getStr()
N = len(S)
ans = ''
at_flag = True
i = 0
while i < N:
if S[i:i+3] == 'dot' and i > 0 and i < N-3:
ans += '.'
i += 3
elif S[i:i+2] == 'at' and at_flag and i > 0:
ans += '@'
at_flag = False
i += 2
else:
ans += S[i]
i += 1
return ans
#for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)á
```
|
instruction
| 0
| 106,517
| 6
| 213,034
|
Yes
|
output
| 1
| 106,517
| 6
| 213,035
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
Submitted Solution:
```
m = input()
m = m.replace("at","@")
m = m.replace("dot",".")
if m[0] == ".":
m = "dot" + m[1:]
if m[-1] == ".":
m = m[:-1] + "dot"
if m[0] == "@":
m = "at" + m[1:]
if m.count("@") > 1:
n = m.find("@")
m = m[:n+1] + m[n+1:].replace("@","at")
print(m)
```
|
instruction
| 0
| 106,518
| 6
| 213,036
|
Yes
|
output
| 1
| 106,518
| 6
| 213,037
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
Submitted Solution:
```
s = str(input())
s = s.replace("at","@")
s = s.replace("dot",".")
if s[0]==".":
s = "dot"+s[1:]
if s[0]=="@":
s = "at"+s[1:]
if s[len(s)-1]==".":
s = s[:len(s)-1]+"dot"
if s[len(s)-1]=="@":
s = s[:len(s)-1]+"at"
print(s)
```
|
instruction
| 0
| 106,520
| 6
| 213,040
|
No
|
output
| 1
| 106,520
| 6
| 213,041
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
Submitted Solution:
```
m = input()
m = m.replace("at","@")
m = m.replace("dot",".")
if m[0] == ".":
m = "dot" + m[1:]
if m[-1] == ".":
m = m[:-1] + "dot"
if m.count("@") > 1:
n = m.find("@")
m = m[:n+1] + m[n+1:].replace("@","at")
print(m)
```
|
instruction
| 0
| 106,521
| 6
| 213,042
|
No
|
output
| 1
| 106,521
| 6
| 213,043
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
Submitted Solution:
```
t = input()
t = t.replace("dot", '.')
t = t.replace("at", '@')
if t.find('.')==0: t = t.replace(".", "dot", 1)
if t.find('@')==0: t = t.replace("@", "at", 1)
if t.rfind('.')==len(t)-1:
t = t[0:len(t)-2]
t += "dot"
if t.rfind('@')==len(t)-1:
t = t[0:len(t)-1]
t += "at"
print(t)
```
|
instruction
| 0
| 106,522
| 6
| 213,044
|
No
|
output
| 1
| 106,522
| 6
| 213,045
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes one has to spell email addresses over the phone. Then one usually pronounces a dot as dot, an at sign as at. As a result, we get something like vasyaatgmaildotcom. Your task is to transform it into a proper email address (vasya@gmail.com).
It is known that a proper email address contains only such symbols as . @ and lower-case Latin letters, doesn't start with and doesn't end with a dot. Also, a proper email address doesn't start with and doesn't end with an at sign. Moreover, an email address contains exactly one such symbol as @, yet may contain any number (possible, zero) of dots.
You have to carry out a series of replacements so that the length of the result was as short as possible and it was a proper email address. If the lengths are equal, you should print the lexicographically minimal result.
Overall, two variants of replacement are possible: dot can be replaced by a dot, at can be replaced by an at.
Input
The first line contains the email address description. It is guaranteed that that is a proper email address with all the dots replaced by dot an the at signs replaced by at. The line is not empty and its length does not exceed 100 symbols.
Output
Print the shortest email address, from which the given line could be made by the described above replacements. If there are several solutions to that problem, print the lexicographically minimal one (the lexicographical comparison of the lines are implemented with an operator < in modern programming languages).
In the ASCII table the symbols go in this order: . @ ab...z
Examples
Input
vasyaatgmaildotcom
Output
vasya@gmail.com
Input
dotdotdotatdotdotat
Output
dot..@..at
Input
aatt
Output
a@t
Submitted Solution:
```
s=input()
i=s.count("at")
j=s.count("dot")
if( i != 1):
index=s.find("at")
index=index+3
s1=s[0:index]
s2=s[index:]
s2=s2.replace("at","@",i-1)
s=s1+s2
else:
s=s.replace("at","@")
if(j !=1):
ind=s.find("dot")
ind=ind+4
s3=s[0:ind]
s4=s[ind:]
s4=s4.replace("dot",".")
s=s3+s4
else:
s=s.replace("dot",".")
print(s)
```
|
instruction
| 0
| 106,523
| 6
| 213,046
|
No
|
output
| 1
| 106,523
| 6
| 213,047
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,153
| 6
| 214,306
|
Tags: implementation, strings
Correct Solution:
```
s=str(input())
if (s[0] == s[0].lower() and s[1:] == s[1:].upper()):
print(s[0].upper()+s[1:].lower())
elif (s == s.upper()):
print(s.lower())
else:
print(s)
```
|
output
| 1
| 107,153
| 6
| 214,307
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,154
| 6
| 214,308
|
Tags: implementation, strings
Correct Solution:
```
a = input()
if a.isupper() or a[1:].isupper() or (len(a)==1 and a.islower()):
a = list(a)
for i in range(len(a)):
if a[i].isupper():
a[i] = a[i].lower()
else:
a[i] = a[i].upper()
print("".join(a))
```
|
output
| 1
| 107,154
| 6
| 214,309
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,155
| 6
| 214,310
|
Tags: implementation, strings
Correct Solution:
```
s=input().strip()
a=s
if len(s)==1:
if a.upper()==s:
print(a.lower())
else:
print(a.upper())
else:
if a.upper()==s:
print(a.lower())
else:
x=a[0].lower()
y=a[1:].upper()
if x==s[0] and y==s[1:]:
x=a[0].upper()
y=a[1:].lower()
print(x+y)
else:
print(s)
```
|
output
| 1
| 107,155
| 6
| 214,311
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,156
| 6
| 214,312
|
Tags: implementation, strings
Correct Solution:
```
s=input()
if s.upper()==s:
print(s.lower())
else:
q=s[0]
i=s[1:]
if s[0]==q.lower() and i.upper()==i:
print(q.upper()+i.lower())
else:
print(s)
```
|
output
| 1
| 107,156
| 6
| 214,313
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,157
| 6
| 214,314
|
Tags: implementation, strings
Correct Solution:
```
s = input()
if len(s)==1:
if( s.islower() ):
#if it is 'z' make it 'Z'
s = s.upper()
else:
#if it is 'Z' make it 'z':
s = s.lower()
elif s[0].islower() and s[1:].isupper():
# hELLO to Hello
s = s.capitalize()
elif s.isupper():
#HELLO --> hello
s = s.lower()
print(s)
```
|
output
| 1
| 107,157
| 6
| 214,315
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,158
| 6
| 214,316
|
Tags: implementation, strings
Correct Solution:
```
word = input()
if len(word)==1 and word.isupper():
print(word.lower())
elif len(word) ==1 and word.islower():
print(word.upper())
else:
if word[1:].isupper() and word[0].islower():
print(word.capitalize())
elif word.isupper():
print(word.lower())
else:
print(word)
```
|
output
| 1
| 107,158
| 6
| 214,317
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,159
| 6
| 214,318
|
Tags: implementation, strings
Correct Solution:
```
n=input()
c=0
if n[0].islower()==True:
c+=1
for i in range(0,len(n)):
if n[i].isupper()==True:
c+=1
r=""
if len(n)==c:
for i in n:
if i.isupper()==True:
r+=i.lower()
else:
r+=i.upper()
print(r)
else:
print(n)
```
|
output
| 1
| 107,159
| 6
| 214,319
|
Provide tags and a correct Python 3 solution for this coding contest problem.
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
* either it only contains uppercase letters;
* or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output
Print the result of the given word's processing.
Examples
Input
cAPS
Output
Caps
Input
Lock
Output
Lock
|
instruction
| 0
| 107,160
| 6
| 214,320
|
Tags: implementation, strings
Correct Solution:
```
#! /usr/bin/env python3
word = input()
if (word.isupper()) or (word[1:].isupper()) or len(word) == 1:
word = word.swapcase()
print(word)
else:
print(word)
```
|
output
| 1
| 107,160
| 6
| 214,321
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo adores strings. But most of all he adores strings of length n.
One day he wanted to find a string that meets the following conditions:
1. The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct.
2. No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2... sn, then the following inequality holds, si ≠ si + 1(1 ≤ i < n).
3. Among all strings that meet points 1 and 2, the required string is lexicographically smallest.
Help him find such string or state that such string doesn't exist.
String x = x1x2... xp is lexicographically less than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there is such number r (r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.
Input
A single line contains two positive integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string's length and the number of distinct letters.
Output
In a single line print the required string. If there isn't such string, print "-1" (without the quotes).
Examples
Input
7 4
Output
ababacd
Input
4 7
Output
-1
|
instruction
| 0
| 107,308
| 6
| 214,616
|
Tags: greedy
Correct Solution:
```
n, k = map(int, input().split())
if k > n or k == 1 and n > 1:
print(-1)
elif k == 1 and n == 1:
print('a')
else:
print(('ab' * (n // 2 + 1))[: (n - (k - 2))] + 'cdefghijklmnopqrstuvwxyz'[:k - 2])
```
|
output
| 1
| 107,308
| 6
| 214,617
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo adores strings. But most of all he adores strings of length n.
One day he wanted to find a string that meets the following conditions:
1. The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct.
2. No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2... sn, then the following inequality holds, si ≠ si + 1(1 ≤ i < n).
3. Among all strings that meet points 1 and 2, the required string is lexicographically smallest.
Help him find such string or state that such string doesn't exist.
String x = x1x2... xp is lexicographically less than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there is such number r (r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.
Input
A single line contains two positive integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string's length and the number of distinct letters.
Output
In a single line print the required string. If there isn't such string, print "-1" (without the quotes).
Examples
Input
7 4
Output
ababacd
Input
4 7
Output
-1
|
instruction
| 0
| 107,310
| 6
| 214,620
|
Tags: greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
n, t = map(int, input().split())
alphas = "abcdefghijklmnopqrstuvwxyz"
if t > n:
print(-1)
elif n > 1 and t == 1:
print(-1)
elif n == 1 and t == 1:
print("a")
elif t == n:
print(alphas[:n])
else:
ans = ''
if (n-(t-2)) % 2 == 1:
ans += "ab"*((n-(t-2))//2)
ans += "a"
if t > 2:
ans += alphas[2:(t-2) + 2]
else:
ans += "ab"*((n-(t-2))//2)
if t > 2:
ans += alphas[2:(t-2) + 2]
print(ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
|
output
| 1
| 107,310
| 6
| 214,621
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo adores strings. But most of all he adores strings of length n.
One day he wanted to find a string that meets the following conditions:
1. The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct.
2. No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2... sn, then the following inequality holds, si ≠ si + 1(1 ≤ i < n).
3. Among all strings that meet points 1 and 2, the required string is lexicographically smallest.
Help him find such string or state that such string doesn't exist.
String x = x1x2... xp is lexicographically less than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there is such number r (r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.
Input
A single line contains two positive integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string's length and the number of distinct letters.
Output
In a single line print the required string. If there isn't such string, print "-1" (without the quotes).
Examples
Input
7 4
Output
ababacd
Input
4 7
Output
-1
|
instruction
| 0
| 107,312
| 6
| 214,624
|
Tags: greedy
Correct Solution:
```
n, k = map(int, input().split())
if n < k or k > 26: print(-1)
elif k == 1: print('a' if n == 1 else -1)
else: print(('ab' * (n // 2 + 1))[: n - k + 2] + 'cdefghijklmnopqrstuvwxyz'[: k - 2])
# Made By Mostafa_Khaled
```
|
output
| 1
| 107,312
| 6
| 214,625
|
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