text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Tags: implementation, sortings, strings
Correct Solution:
```
def solve(s):
n = r = 0
for c in s:
if c == "n":
n += 1
continue
if c == "r":
r += 1
ans = " ".join(["1" for i in range(n)] + ["0" for i in range(r)])
return ans
input()
s = input()
print(solve(s))
```
| 103,000 |
Provide tags and a correct Python 3 solution for this coding contest problem.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Tags: implementation, sortings, strings
Correct Solution:
```
length = int(input())
string = input()
contador_0 = 0
contador_1 = 0
for letter in string:
if letter == 'z':
contador_0 += 1
elif letter == 'n':
contador_1 += 1
string = ''
for contador in range(contador_1):
string += '1 '
for contador in range(contador_0):
string += '0 '
string=''.join(list(string)[:-1])
print(string)
```
| 103,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
n=int(input())
s=input()
z=s.count('z')
o=n-z*4
l=[1]*(o//3)
l+=[0]*z
print(*l)
```
Yes
| 103,002 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
d = {
'z':0,
'e':0,
'r':0,
'o':0,
'n':0,
'e':0
}
input()
for i in input():
d[i] += 1
one = 0
null = 0
m = min(d['o'], d['n'], d['e'])
if m <= d['o'] and d['n'] >= m and d['e'] >= m:
one += m
d['e'] -= m
d['o'] -= m
m = min(d['z'], d['e'], d['r'], d['o'])
if m >= d['z'] and m >= d['e'] and m >= d['r'] and m >= d['o']:
null += m
print('1 ' * one, '0 ' * null)
```
Yes
| 103,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Fri Nov 6 00:39:35 2020
@author: Xuan Loc
"""
if __name__=='__main__':
n=int(input())
s=list(input())
my_dict={'z':0,'e':0,'r':0,'o':0,'n':0}
for x in s:
my_dict[x]+=1
one=min(my_dict['o'],my_dict['n'],my_dict['e'])
zero=min(my_dict['z'],my_dict['r'],my_dict['e']-one,my_dict['o']-one)
for i in range(one):
print(1,end=' ')
for i in range(zero):
print(0,end=' ')
```
Yes
| 103,004 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
n = int(input())
s = input()
print(' '.join(list('1'*s.count('n')+'0' * s.count('z'))))
```
Yes
| 103,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
a = int(input())
b = str(input())
g =[]
for i in range(0, len(b)-1):
if b[i] == "z":
g.append(0)
if b[i] == "n":
g.append(1)
print(*g)
```
No
| 103,006 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
n = int(input())
cards = list(input())
zeros = cards.count("z")
ones = cards.count("o") - zeros
bn = []
for i in range(ones):
bn.append("1")
for i in range(zeros):
bn.append("0")
print("".join(bn))
```
No
| 103,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
n=int(input())
s=input()
d={"n":0,"z":0}
for i in s:
if i in d:
d[i]+=1
x,y=d["n"],d["z"]
l="1"*x+"0"*y
l=int(l)
print(l)
```
No
| 103,008 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input
The first line contains a single integer n (1 ⩽ n ⩽ 10^5) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 0 or "one" which corresponds to the digit 1.
Output
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Examples
Input
4
ezor
Output
0
Input
10
nznooeeoer
Output
1 1 0
Note
In the first example, the correct initial ordering is "zero".
In the second example, the correct initial ordering is "oneonezero".
Submitted Solution:
```
s = input()
n = s.count('n')
r = s.count('r')
for i in range(n):
print(1,end=' ')
for i in range(r):
print(0,end=' ')
```
No
| 103,009 |
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'.
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)
```
| 103,010 |
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'.
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)
```
| 103,011 |
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'.
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)
```
| 103,012 |
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'.
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)
```
| 103,013 |
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'.
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)
```
| 103,014 |
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'.
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()
```
| 103,015 |
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'.
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)])
```
| 103,016 |
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'.
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])
```
| 103,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
# sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
def YES(c): return IF(c, "YES", "NO")
def Yes(c): return IF(c, "Yes", "No")
def main():
t = 1
rr = []
for _ in range(t):
s = S()
dp = [1,0,0]
for c in s:
np = [0] * 3
t = sum(dp) % mod
if c == 'u':
np[0] = dp[1]
np[1] = t
elif c == 'n':
np[0] = dp[2]
np[2] = t
elif c in 'mw':
dp = [0]
break
else:
np[0] = t
dp = np
rr.append(sum(dp) % mod)
return JA(rr, "\n")
print(main())
```
Yes
| 103,018 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
MOD = 1000000007
ii = lambda: int(input())
si = lambda: input()
dgl = lambda: list(map(int, input()))
f = lambda: list(map(int, input().split()))
il = lambda: list(map(int, input().split()))
ls = lambda: list(input())
s=si()
n=len(s)
dp=[0]*(n+2)
dp[0]=1
dp[1]=1 if not s[0] in ['m','w'] else 0
for i in range(2,n+1):
if s[i-1] in ['m','w']:
exit(print(0))
dp[i]=dp[i-1]
if s[i-1]=='n' or s[i-1]=='u':
if s[i-2]==s[i-1]:
dp[i]+=dp[i-2]
dp[i]%=MOD
print(dp[n])
```
Yes
| 103,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
n = input()
count = 1
for i in n:
if i in ['w', 'm']:
print(0)
exit()
dp = [0] * (len(n) + 1)
dp[0] = 1
dp[1] = 1
for i in range(2, len(n) + 1):
dp[i] = dp[i - 1]
if n[i - 1] == n[i - 2] and n[i - 1] in ['u', 'n']:
dp[i] = (dp[i] + dp[i - 2]) % 1000000007
print(dp[len(n)])
```
Yes
| 103,020 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
import math
import bisect
import copy
from collections import defaultdict
from collections import OrderedDict
#for _ in range(int(input())):
#n = int(input())
s = list(input())
n = len(s)
#print(s)
#a,b,n = map(int,input().split())
#ar = list(map(int,input().split()))
#br = list(map(int,input().split()))
c = []
i = 0
x = 1
while i < n :
while i < n-1 and s[i] == s[i+1] == 'u':
x += 1
i += 1
if i == n-1 or s[i] != s[i+1]:
break
if x>1:
c.append(x)
x = 1
while i < n-1 and s[i] == s[i+1] == 'n':
x += 1
i += 1
if i == n-1 or s[i] != s[i+1]:
break
if x>1:
c.append(x)
x = 1
if x == 1:
i += 1
fibo = [1,2]
#print(c,fibo)
ans = 0
if n == 1:
if s[0] == 'w' or s[0] == 'm':
ans = -1
else:
for i in range(n):
if i== 0:
if s[i] =='w' and s[i] != s[i+1]:
ans = -1
break
elif s[i] == 'm' and s[i] != s[i+1]:
ans = -1
break
elif i == n-1:
if s[i] =='w' and s[i] != s[i-1]:
ans = -1
break
elif s[i] == 'm' and s[i] != s[i-1]:
ans = -1
break
else:
if s[i] == 'w' and s[i-1] != 'w' and s[i+1] != 'w':
ans = -1
break
elif s[i] == 'm' and s[i-1]!= 'm' and s[i-1] != 'm':
ans = -1
break
if ans == -1:
print(0)
else:
if len(c) == 0:
print(1)
else:
for i in range(2,max(c)):
fibo.append((fibo[i-2] + fibo[i-1])%int(1e9 + 7))
ans = 1
for j in range(len(c)):
ans *= fibo[c[j]-1]
ans %= int(1e9 + 7)
print(ans)
"""
stuff you should look for
* int overflow, array bounds
* special cases (n=1?)
* do smth instead of nothing and stay organized
* WRITE STUFF DOWN
* DON'T GET STUCK ON ONE APPROACH
"""
```
Yes
| 103,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
mas = [0,1,2]
b = 0
for i in range(3,10**5 + 1):
mas.append((mas[i-1] + mas[i-2])%(10**9 + 7))
ans = 1
s = input()
f = -1
for i in range(len(s)):
if i > f:
if i < len(s) and s[i] == 'n':
lans = 0
while i < len(s) and s[i] == 'n':
lans += 1
i += 1
ans *= mas[lans]
ans = ans % (10**9 + 7)
if i < len(s) and s[i] == 'u':
lans = 0
while i < len(s) and s[i] == 'u':
lans += 1
i += 1
ans *= mas[lans]
ans = ans % (10**9 + 7)
if i < len(s) and (s[i] == 'm' or s[i] == 'w'):
b = 1
break
f = i
if b == 1:
print(0)
else:
print(ans % (10**9 + 7))
```
No
| 103,022 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
from sys import stdin,stdout
mod=10**9 + 7
s=input().rstrip()
s=[ i for i in s]
n=len(s)
dp=[1]*(n)
if 'm' in s or 'w' in s:
print(0)
exit()
if len(s)==2:
if s[0]=='u' and s[1]=='u':
print(2)
elif s[0]=='n' and s[0]=='n':
print(2)
else:
print(1)
exit()
dp[0]=0
if len(s)>2:
if s[0]==s[1] and (s[0]=='u' or s[0]=='n'):
dp[0]=1
for i in range(2,len(s)):
dp[i]=dp[i-1]
if s[i]=='n' and s[i-1]=='n':
dp[i]=max(dp[i-2]+2,dp[i])
if s[i]=='u' and s[i-1]=='u':
dp[i]=max(dp[i-2]+2,dp[i])
print(dp[n-1]%mod)
```
No
| 103,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
s=input()
dp=[0]*(len(s))
dp[0]=1
temp=0
for i in range(1,len(s)):
if(s[i]=='w' or s[i]=='m'):
temp=1
break
elif((s[i]=='u' and s[i-1]=='u') or (s[i]=='n' and s[i-1]=='n')):
if(i!=1):
dp[i]=dp[i-1]+dp[i-2]
else:
dp[i]=dp[i-1]+1
else:
dp[i]=dp[i-1]
# print(dp)
if(temp==1):
print(0)
else:
print(dp[len(s)-1])
```
No
| 103,024 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
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'.
Submitted Solution:
```
S = input()
N = len(S)
out = 1
mod = (10**9)+7
"""def f(num):
if num%6 == 1:
return 1
elif num%6 == 2:
return 2
elif num%6 == 3:
return 1
elif num%6 == 4:
return (10**9)+6
elif num%6 == 5:
return (10**9) +5
else:
return (10**9)+6"""
f = [0 for a in range(10**5)]
f[1] = 1
f[2] = 2
for b in range(3, (10**5)):
f[b] = ((f[b-1]%mod)+(f[b-2]%mod))%mod
for i in range(N):
if i < N and S[i] == "u":
c = 1
i += 1
while i < N and S[i] == "u":
c += 1
i += 1
out = ((out%mod)*(f[c]%mod))%mod
if i == N:
break
if i < N and S[i] == "n":
c = 1
i += 1
while i < N and S[i] == "n":
c += 1
i += 1
out = ((out%mod)*(f[c]%mod))%mod
if i == N:
break
if i < N and (S[i] == "m" or S[i] == "w"):
out = 0
break
print(out)
```
No
| 103,025 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
d = {'a', 'b', 'c'}
for _ in range(int(input())):
s = input()
if s.find('aa') != -1 or s.find('bb') != -1 or s.find('cc') != -1:
print(-1)
elif s == '?':
print('a')
else:
s = list(s)
if s[0] == '?':
if s[1] in d:
for c in d:
if c != s[1]:
s[0] = c
break
else:
s[0] = 'a'
if s[-1] == '?':
if s[-2] in d:
for c in d:
if c != s[-2]:
s[-1] = c
break
else:
s[-1] = 'a'
for i in range(1, len(s) - 1):
if s[i] == '?':
if s[i + 1] == '?':
for c in d:
if c != s[i - 1]:
s[i] = c
else:
s[i] = (d ^ set([s[i - 1], s[i + 1]])).pop()
print(''.join(s))
```
| 103,026 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
import collections
from collections import Counter
import itertools
import math
import timeit
#input = sys.stdin.readline
#########################
# imgur.com/Pkt7iIf.png #
#########################
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=1):
divisors = []
for i in range(start, int(math.sqrt(n) + 1)):
if n % i == 0:
if n / i == i:
divisors.append(i)
else:
divisors.extend([i, n // i])
return divisors
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def flin(d, x, default = -1):
left = right = -1
for i in range(len(d)):
if d[i] == x:
if left == -1: left = i
right = i
if left == -1:
return (default, default)
else:
return (left, right)
def ceil(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def prr(a, sep=' '): print(sep.join(map(str, a)))
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
t = ii()
for _ in range(t):
s = f'?{input()}?'
s = [x for x in s]
f = 1
for i in range(1, len(s) - 1):
if s[i] != '?' and (s[i - 1] == s[i] or s[i] == s[i + 1]):
f = 0
break
elif s[i] == '?':
pojiloi_setik = {'a', 'b', 'c'} - {s[i - 1], s[i + 1]}
s[i] = pojiloi_setik.pop()
prr(s[1:-1], '') if f else print(-1)
```
| 103,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for i in range(t):
st=input()
ans=''
fl = False
for i in range(len(st)-1):
if st[i]!="?":
ans+=st[i]
if i!=0:
if st[i-1]==st[i]:
fl = True
break
else:
if i>=1:
if ans[i-1]=="a":
if st[i+1]!="b":
ans+="b"
else:
ans+="c"
elif ans[i-1]=="b":
if st[i+1]!="a":
ans+="a"
else:
ans+="c"
elif ans[i-1]=="c":
if st[i+1]!="a":
ans+="a"
else:
ans+="b"
else:
if st[i+1]=="a":
ans+="b"
elif st[i+1]=="b":
ans+="c"
else:
ans+="a"
if len(st)>1 and st[-1]==st[-2] and st[-1]!="?":
fl=True
if fl:
print(-1)
continue
if st[-1]=="?":
if len(st)==1:
print("a")
else:
if ans[-1]=="a":
ans+="b"
elif ans[-1]=="b":
ans+="c"
else:
ans+="a"
print(ans)
else:
ans+=st[-1]
print(ans)
```
| 103,028 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
lis=['a','b','c']
for _ in range(int(input())):
s=list(input())
f,i=0,0
if(len(s)==1 and s[0]=='?'):
print('a')
continue
if(s[0]=='?'):
for k in range(0,3):
if(s[1]!=lis[k]):
s[0]=lis[k]
else:
i=i+1
if(len(s)==1):
ss=''
print(ss.join(s))
continue
i=0
for j in range(0,len(s)-1):
if(s[j]==s[j+1] and s[j]!='?' and s[j+1]!='?'):
print(-1)
f=1
break
elif(s[j]=='?'):
i=0
for k in range(0,3):
if(lis[k]!=s[(j-1)] and lis[k]!=s[j+1]):
s[j]=lis[k]
else:
i=i+1
if(f==1):
f=0
continue
else:
if(s[len(s)-1]=='?'):
i=0
for k in range(0,3):
if(s[len(s)-2]!=lis[k]):
s[len(s)-1]=lis[k]
else:
i=i+1
ss=""
print(ss.join(s))
```
| 103,029 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
t = int(input())
while t:
t += -1
s = input()
s = list(s)
s.insert(0, '0')
s.append('0')
s.append('1')
ch = 1
for i in range(1, len(s) - 1):
if s[i] == '?':
if s[i - 1] != 'a' and s[i + 1] != 'a': s[i] = 'a'
elif s[i - 1] != 'b' and s[i + 1] != 'b': s[i] = 'b'
else: s[i] = 'c'
if s[i] == s[i + 1]:
ch = 0
break
if ch == 0: print(-1)
else:
s = ''.join(s)
print(s[1: len(s) - 2])
```
| 103,030 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for i in range(t):
s=input()
if "aa" in s or "bb" in s or "cc" in s:
print(-1)
elif len(s)==1 and s[0]=="?":
print("a")
else:
s=list(s)
for j in range(len(s)):
if s[j]=="?":
if j!=0 and j!=len(s)-1:
if s[j-1]!="a" and s[j+1]!="a":s[j]="a"
elif s[j-1]!="b" and s[j+1]!="b":s[j]="b"
elif s[j-1]!="c" and s[j+1]!="c":s[j]="c"
elif j==0:
if s[j+1]!="a":s[j]="a"
elif s[j+1]!="b":s[j]="b"
elif s[j+1]!="c":s[j]="c"
elif j==len(s)-1:
if s[j-1]!="a":s[j]="a"
elif s[j-1]!="b":s[j]="b"
elif s[j-1]!="c":s[j]="c"
print(*s,sep="")
```
| 103,031 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
#code
def solution(a,b):
s=['a','b','c']
s=set(s)
t=[a,b]
s1= s-set(t)
s1=list(s1)
return (s1[0])
def checkString(s):
if(len(s)==1):
return "a"
for i in range(len(s)):
if(s[i]=='?'):
if(i==0):
if(s[i+1]=='?'):
s[i]='a'
else:
s[i]=solution(s[i+1],'a')
elif(i==len(s)-1):
s[i]=solution(s[i-1],'a')
elif(s[i+1]=='?'):
s[i]=solution(s[i-1],'a')
else:
s[i]=solution(s[i-1],s[i+1])
else:
if(i+1 != len(s) and s[i]==s[i+1]):
return -1
return s
# print(s)
for _ in range(int(input())):
s=input()
st=checkString(list(s))
if(st==-1):
print(-1)
else:
print(''.join(st))
```
| 103,032 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Tags: constructive algorithms, greedy
Correct Solution:
```
def dryrun(s):
for i in range(1,len(s)):
if(s[i]!='?'):
if(s[i-1]==s[i]):
return 0
return 1
def replaceq(l,r):
re = ''
if(l==0 and r==0):
re+='a'
elif(l!=0 and r!=0):
if(l==r=='a'):
re+='b'
elif(l==r=='b'):
re+='c'
elif(l==r=='c'):
re+='a'
elif((l=='c' and r=='b') or (l=='b' and r=='c')):
re+='a'
elif((l=='a' and r=='b') or (l=='b' and r=='a')):
re+='c'
elif((l=='a' and r=='c') or (l=='c' and r=='a')):
re+='b'
elif(l==0):
if(r=='a'):
re+='b'
elif(r=='b'):
re+='c'
elif(r=='c'):
re+='a'
elif(r==0):
if(l=='a'):
re+='b'
elif(l=='b'):
re+='c'
elif(l=='c'):
re+='a'
return re
t=int(input())
for _ in range(t):
s = input()
ns = []
if(dryrun(s)):
for i in range(len(s)):
if(s[i]=="?"):
l = 0 if i==0 else ns[i-1]
r = 0 if (i==len(s)-1 or s[i+1]=="?") else s[i+1]
ns.append(replaceq(l,r))
else:
ns.append(s[i])
print(''.join(ns))
else:
print(-1)
```
| 103,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
t=int(input())
for q in range(t):
r={'a':'b','b':'c','c':'a'}
s=list(input())
n=len(s)
c=0
flag=0
for i in range(0,n):
if s[i]!='?':
if i>=1 and s[i]==s[i-1]:
flag=1
break
elif s[i]=='?':
if i==0:
s[i]='a'
else:
s[i]=r[s[i-1]]
if i<n-1:
if s[i]==s[i+1]:
s[i]=r[s[i]]
if flag==1:
print(-1)
else:
print("".join(s))
```
Yes
| 103,034 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
t = int(input())
all = ['a', 'b', 'c']
for _ in range(t):
s = input()
out = []
for i in range(len(s) - 1):
if s[i] != '?' and s[i] == s[i + 1]:
print(-1)
break
else:
if len(s) > 1:
for i in range(len(s)):
ans = ''
if i == 0:
if s[i] == '?':
ans = 'a' if s[i + 1] != 'a' else 'b'
elif i == len(s) - 1:
if s[i] == '?':
ans = 'a' if out[i - 1] != 'a' else 'b'
else:
if s[i] == '?':
ans = list(set(all) - set([out[i-1], s[i+1]]))[0]
if ans == '':
out.append(s[i])
else:
out.append(ans)
else:
out = ['a']
print(''.join(out))
```
Yes
| 103,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
t=int(input())
for x in range(t):
a=list(input())
if len(a)>=2 and a[0]=='?':
if a[1]=='a':
a[0]='b'
elif a[1]=='b':
a[0]='a'
elif a[1]=='c':
a[0]='b'
elif a[1]=='?':
a[0]='a'
elif a[0]=='?':
a[0]='a'
for x in range(1,len(a)-1):
if a[x]=='?':
g=a[x-1]+a[x+1]
if g=='ab' or g=='ba':
a[x]='c'
elif g=='ac' or g=='ca':
a[x]='b'
elif g=='bc' or g=='cb':
a[x]='a'
if g=='aa':
a[x]='b'
elif g=='bb':
a[x]='a'
elif g=='cc':
a[x]='b'
if g=='a?':
a[x]='b'
elif g=='b?':
a[x]='a'
elif g=='c?':
a[x]='b'
if len(a)>=2 and a[-1]=='?':
if a[-2]=='a':
a[-1]='b'
elif a[-2]=='b':
a[-1]='a'
elif a[-2]=='c':
a[-1]='b'
elif a[-2]=='?':
a[-1]='a'
elif a[-1]=='?':
a[-1]='a'
t=True
for x in range(len(a)-1):
if a[x]==a[x+1]:
t=False
if t:print(''.join(a))
else:print(-1)
```
Yes
| 103,036 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
def abc(s: str):
if not s:
return s
l = len(s)
if l == 1:
if s == '?':
return 'a'
else:
return s
s = list(s)
for i in range(l-1):
if s[i] == s[i+1] and s[i] != '?':
return '-1'
for i in range(l):
if s[i] != '?':
continue
if i == 0:
if s[1] == '?':
s[0] = 'a'
elif s[1] == 'a':
s[0] = 'b'
else:
s[0] = 'a'
continue
if i == l-1:
if s[-2] == '?':
s[-1] = 'a'
elif s[-2] == 'a':
s[-1] = 'b'
else:
s[-1] = 'a'
continue
for j in ['a', 'b', 'c']:
if j != s[i-1] and j != s[i+1]:
s[i] = j
break
return ''.join(s)
n = int(input())
out = []
for i in range(n):
s = input()
out.append(abc(s))
for i in out:
print(i)
```
Yes
| 103,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
n=int(input())
for i in range(n):
s=str(input())
s=list(s)
if(len(s)==1):
if(s[0]=='?'):
print('a')
else:
print(s[0])
else:
flag=0
for j in range(0,len(s)-1):
if(s[j]!='?' and s[j]==s[j+1]):
flag=1
break
if(flag==1):
print(-1)
else:
e=''
for k in range(0,len(s)):
if(s[k]!='?'):
e=e+s[k]
elif(s[k]=='?' and k==0):
if(s[k+1]=='a'):
e=e+'b'
elif(s[k+1]=='b'):
e=e+'c'
else:
e=e+'a'
elif(s[k]=='?' and k==len(s)-1):
if(s[k-1]=='a'):
e=e+'b'
elif(s[k-1]=='b'):
e=e+'c'
else:
e=e+'a'
else:
l=[]
l.append(s[k+1])
l.append(s[k-1])
if('a' not in l):
e=e+'a'
s[k]='a'
elif('b' not in l):
e=e+'b'
s[k]='b'
elif('c' not in l):
e=e+'c'
s[k]='c'
print(e)
```
No
| 103,038 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
import random
t = int(input())
for _ in range(t):
txt = input()
for i in range(len(txt)-1):
chars = ['a', 'b', 'c']
if txt[i] != '?' and txt[i] == txt[i+1]:
print(-1)
break
if txt[i] == '?':
if i>0 and txt[i-1] != '?':
chars.remove(txt[i-1])
if i<(len(txt)-1) and txt[i+1] != '?' and txt[i+1] in chars:
chars.remove(txt[i+1])
k = random.randint(0, len(chars)-1)
txt = txt[:i]+chars[k]+txt[i+1:]
# txt[i] = chars[k]
else:
print(txt)
```
No
| 103,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
def getChar(x, y):
opts = ['a', 'b', 'c']
if x in opts: opts.remove(x)
if y in opts: opts.remove(y)
return opts[0]
def findBS(s):
length = len(s)
opts = ['a', 'b', 'c']
if length == 1:
if '?' in s: print(-1)
else: print(s)
return
for i in range(len(opts)):
if s[0] == '?' and s[1] != opts[i]:
s[0] = opts[i]
if s[length-1] == '?' and s[length-2] != opts[i]:
s[length-1] = opts[i]
for i in range(1, length-1):
if s[i] == '?': s[i] = getChar(s[i-1], s[i+1])
if s[i] == s[i-1] or s[i] == s[i+1]:
print('-1')
return
if '?' in opts:
print(-1)
print(''.join(s))
t = int(input())
for _ in range(t):
s = list(input().rstrip())
findBS(s)
```
No
| 103,040 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not.
Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him!
More formally, after replacing all characters '?', the condition s_i ≠ s_{i+1} should be satisfied for all 1 ≤ i ≤ |s| - 1, where |s| is the length of the string s.
Input
The first line contains positive integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain the descriptions of test cases.
Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'.
It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5.
Output
For each test case given in the input print the answer in the following format:
* If it is impossible to create a beautiful string, print "-1" (without quotes);
* Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful.
In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.
In the third test case, the only answer is "acbac".
Submitted Solution:
```
def solve(p):
if not p:
return []
if len(p)==1:
if p[0] == '?':
return ['a']
else:
return p
s = set()
s.add('a')
s.add('b')
s.add('c')
for i in range(1,len(p)-1):
if p[i] == '?':
rem = list(s.difference(set([p[i-1], p[i+1], '?'])))
if not rem:
return -1
else:
p[i] = rem[0]
if p[0] == '?':
p[0] = list(s.difference(set([p[1], '?'])))[0]
if p[-1] == '?':
p[-1] = list(s.difference(set([p[-2], '?'])))[0]
for i in range(1, len(p)):
if p[i]==p[i-1]:
return -1
return p
problems = []
for i in range(int(input())):
problems += [list(input())]
for p in problems:
print(p)
p = solve(p)
if p!= -1:
p = ''.join(p)
print(p)
```
No
| 103,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different with easy version only by constraints on total answers length
It is an interactive problem
Venya joined a tour to the madhouse, in which orderlies play with patients the following game. Orderlies pick a string s of length n, consisting only of lowercase English letters. The player can ask two types of queries:
* ? l r – ask to list all substrings of s[l..r]. Substrings will be returned in random order, and in every substring, all characters will be randomly shuffled.
* ! s – guess the string picked by the orderlies. This query can be asked exactly once, after that the game will finish. If the string is guessed correctly, the player wins, otherwise he loses.
The player can ask no more than 3 queries of the first type.
To make it easier for the orderlies, there is an additional limitation: the total number of returned substrings in all queries of the first type must not exceed \left⌈ 0.777(n+1)^2 \right⌉ (⌈ x ⌉ is x rounded up).
Venya asked you to write a program, which will guess the string by interacting with the orderlies' program and acting by the game's rules.
Your program should immediately terminate after guessing the string using a query of the second type. In case your program guessed the string incorrectly, or it violated the game rules, it will receive verdict Wrong answer.
Note that in every test case the string is fixed beforehand and will not change during the game, which means that the interactor is not adaptive.
Input
First line contains number n (1 ≤ n ≤ 100) — the length of the picked string.
Interaction
You start the interaction by reading the number n.
To ask a query about a substring from l to r inclusively (1 ≤ l ≤ r ≤ n), you should output
? l r
on a separate line. After this, all substrings of s[l..r] will be returned in random order, each substring exactly once. In every returned substring all characters will be randomly shuffled.
In the case, if you ask an incorrect query, ask more than 3 queries of the first type or there will be more than \left⌈ 0.777(n+1)^2 \right⌉ substrings returned in total, you will receive verdict Wrong answer.
To guess the string s, you should output
! s
on a separate line.
After printing each query, do not forget to flush the output. Otherwise, you will get Idleness limit exceeded. To flush the output, you can use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you received - (dash) as an answer to any query, you need to terminate your program with exit code 0 (for example, by calling exit(0)). This means that there was an error in the interaction protocol. If you don't terminate with exit code 0, you can receive any unsuccessful verdict.
Hack format
To hack a solution, use the following format:
The first line should contain one integer n (1 ≤ n ≤ 100) — the length of the string, and the following line should contain the string s.
Example
Input
4
a
aa
a
cb
b
c
c
Output
? 1 2
? 3 4
? 4 4
! aabc
Submitted Solution:
```
from sys import stdout
class String2:
def __init__(self, s, b=None):
if b is not None:
self.b = b
return
self.b = [0] * 26
for i in s:
self.b[ord(i) - ord('a')] += 1
def __add__(self, other):
b = self.b.copy()
for i in range(26):
b[i] += other.b[i]
return String2('', b)
def __sub__(self, other):
b = self.b.copy()
for i in range(26):
b[i] -= other.b[i]
return b
def __mul__(self, other):
ans = String2('', self.b)
for i in range(26):
ans.b[i] *= other
return ans
def req(l, r, k=0):
print('?', l, r)
v = [''.join(sorted(input())) for i in range((r - l + 1) * (r - l + 2) // 2)]
stdout.flush()
return v
def compute(v):
bukvi = [[0] * (n + 2) for _ in range(26)]
for el in v:
cur = len(el)
for e in el:
bukvi[ord(e) - ord('a')][cur] += 1
return bukvi
def compute2(bukvi):
bukvis = [set() for i in range(n + 2)]
for i in range(26):
prev = bukvi[i][1]
for j in range(1, n // 2 + n % 2 + 1):
while bukvi[i][j] != prev:
bukvis[j].add(chr(ord('a') + i))
prev += 1
return bukvis
def solve(va, va2):
for i in va2:
va.remove(i)
va.sort(key=len)
s = va[0]
for i in range(1, len(va)):
for j in range(26):
if va[i].count(chr(ord('a') + j)) != va[i - 1].count(chr(ord('a') + j)):
s += chr(ord('a') + j)
return s
def check(v, s, s2, f):
s3 = String2(v[0])
for i in range(1, len(v)):
s3 = s3 + String2(v[i])
le = len(v[0])
cur = String2('', String2('', f - String2(s)) - String2(s2)) * le
for i in range(le - 2):
cur = cur + (String2(s[i]) * (i + 1)) + (String2(s2[-i]) * (i + 1))
cur = cur + (String2(s[le - 2]) * (le - 1)) + (String2(s[le - 1]) * le)
e = cur - s3
for i in range(26):
if e[i]:
return chr(ord('a') + i)
def main():
va = req(1, n)
va2 = req(1, n // 2 + n % 2)
va3 = req(2, n // 2 + n % 2)
# bukvi2 = compute(va2)
# bukvi3 = compute(va3)
if n == 1:
print('!', va[0][0])
return
ma = [[] for i in range(n * 2)]
for i in va:
ma[len(i)].append(i)
a = String2(''.join(ma[1]))
s = solve(va2, va3)
s2 = ''
for i in range(2, n // 2 + 1):
s2 = check(ma[i], s, s2, a) + s2
se = String2('', a - String2(s)) - String2(s2)
for i in range(len(se)):
if se[i]:
s += chr(ord('a') + i)
break
print(s + s2)
n = int(input())
main()
```
No
| 103,042 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different with easy version only by constraints on total answers length
It is an interactive problem
Venya joined a tour to the madhouse, in which orderlies play with patients the following game. Orderlies pick a string s of length n, consisting only of lowercase English letters. The player can ask two types of queries:
* ? l r – ask to list all substrings of s[l..r]. Substrings will be returned in random order, and in every substring, all characters will be randomly shuffled.
* ! s – guess the string picked by the orderlies. This query can be asked exactly once, after that the game will finish. If the string is guessed correctly, the player wins, otherwise he loses.
The player can ask no more than 3 queries of the first type.
To make it easier for the orderlies, there is an additional limitation: the total number of returned substrings in all queries of the first type must not exceed \left⌈ 0.777(n+1)^2 \right⌉ (⌈ x ⌉ is x rounded up).
Venya asked you to write a program, which will guess the string by interacting with the orderlies' program and acting by the game's rules.
Your program should immediately terminate after guessing the string using a query of the second type. In case your program guessed the string incorrectly, or it violated the game rules, it will receive verdict Wrong answer.
Note that in every test case the string is fixed beforehand and will not change during the game, which means that the interactor is not adaptive.
Input
First line contains number n (1 ≤ n ≤ 100) — the length of the picked string.
Interaction
You start the interaction by reading the number n.
To ask a query about a substring from l to r inclusively (1 ≤ l ≤ r ≤ n), you should output
? l r
on a separate line. After this, all substrings of s[l..r] will be returned in random order, each substring exactly once. In every returned substring all characters will be randomly shuffled.
In the case, if you ask an incorrect query, ask more than 3 queries of the first type or there will be more than \left⌈ 0.777(n+1)^2 \right⌉ substrings returned in total, you will receive verdict Wrong answer.
To guess the string s, you should output
! s
on a separate line.
After printing each query, do not forget to flush the output. Otherwise, you will get Idleness limit exceeded. To flush the output, you can use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you received - (dash) as an answer to any query, you need to terminate your program with exit code 0 (for example, by calling exit(0)). This means that there was an error in the interaction protocol. If you don't terminate with exit code 0, you can receive any unsuccessful verdict.
Hack format
To hack a solution, use the following format:
The first line should contain one integer n (1 ≤ n ≤ 100) — the length of the string, and the following line should contain the string s.
Example
Input
4
a
aa
a
cb
b
c
c
Output
? 1 2
? 3 4
? 4 4
! aabc
Submitted Solution:
```
from sys import stdout
class String2:
def __init__(self, s, b=None):
if b is not None:
self.b = b
return
self.b = [0] * 26
for i in s:
self.b[ord(i) - ord('a')] += 1
def __add__(self, other):
b = self.b.copy()
for i in range(26):
b[i] += other.b[i]
return String2('', b)
def __sub__(self, other):
b = self.b.copy()
for i in range(26):
b[i] -= other.b[i]
return b
def __mul__(self, other):
ans = String2('', self.b)
for i in range(26):
ans.b[i] *= other
return ans
def req(l, r, k=0):
print('?', l, r)
v = [''.join(sorted(input())) for i in range((r - l + 1) * (r - l + 2) // 2)]
stdout.flush()
return v
def compute(v):
bukvi = [[0] * (n + 2) for _ in range(26)]
for el in v:
cur = len(el)
for e in el:
bukvi[ord(e) - ord('a')][cur] += 1
return bukvi
def compute2(bukvi):
bukvis = [set() for i in range(n + 2)]
for i in range(26):
prev = bukvi[i][1]
for j in range(1, n // 2 + n % 2 + 1):
while bukvi[i][j] != prev:
bukvis[j].add(chr(ord('a') + i))
prev += 1
return bukvis
def solve(va, va2):
for i in va2:
va.remove(i)
va.sort(key=len)
s = va[0]
for i in range(1, len(va)):
for j in range(26):
if va[i].count(chr(ord('a') + j)) != va[i - 1].count(chr(ord('a') + j)):
s += chr(ord('a') + j)
return s
def check(v, s, s2, f):
s3 = String2(v[0])
for i in range(1, len(v)):
s3 = s3 + String2(v[i])
le = len(v[0])
cur = String2('', String2('', f - String2(s)) - String2(s2)) * le
for i in range(le - 2):
cur = cur + (String2(s[i]) * (i + 1)) + (String2(s2[-i]) * (i + 1))
cur = cur + (String2(s[le - 2]) * (le - 1)) + (String2(s[le - 1]) * le)
e = cur - s3
for i in range(26):
if e[i]:
return chr(ord('a') + i)
def main():
if n == 1:
va = req(1, 1)
print('!', va[0])
return
elif n == 2:
va2 = req(1, 1)
va3 = req(2, 2)
print('!', va2[0] + va3[0])
return
elif n == 3:
va = req(1, 1)
va2 = req(2, 2)
va3 = req(3, 3)
print('!', va[0] + va2[0] + va3[0])
return
va = req(1, n)
va2 = req(1, max(n // 2 + n % 2, 2))
va3 = req(2, max(n // 2 + n % 2, 2))
# bukvi2 = compute(va2)
# bukvi3 = compute(va3)
ma = [[] for i in range(n * 2)]
for i in va:
ma[len(i)].append(i)
a = String2(''.join(ma[1]))
s = solve(va2, va3)
s2 = ''
for i in range(2, n // 2 + 1):
s2 = check(ma[i], s, s2, a) + s2
se = String2('', a - String2(s)) - String2(s2)
for i in range(len(se)):
if se[i]:
s += chr(ord('a') + i)
break
print('!', s + s2)
n = int(input())
main()
```
No
| 103,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different with easy version only by constraints on total answers length
It is an interactive problem
Venya joined a tour to the madhouse, in which orderlies play with patients the following game. Orderlies pick a string s of length n, consisting only of lowercase English letters. The player can ask two types of queries:
* ? l r – ask to list all substrings of s[l..r]. Substrings will be returned in random order, and in every substring, all characters will be randomly shuffled.
* ! s – guess the string picked by the orderlies. This query can be asked exactly once, after that the game will finish. If the string is guessed correctly, the player wins, otherwise he loses.
The player can ask no more than 3 queries of the first type.
To make it easier for the orderlies, there is an additional limitation: the total number of returned substrings in all queries of the first type must not exceed \left⌈ 0.777(n+1)^2 \right⌉ (⌈ x ⌉ is x rounded up).
Venya asked you to write a program, which will guess the string by interacting with the orderlies' program and acting by the game's rules.
Your program should immediately terminate after guessing the string using a query of the second type. In case your program guessed the string incorrectly, or it violated the game rules, it will receive verdict Wrong answer.
Note that in every test case the string is fixed beforehand and will not change during the game, which means that the interactor is not adaptive.
Input
First line contains number n (1 ≤ n ≤ 100) — the length of the picked string.
Interaction
You start the interaction by reading the number n.
To ask a query about a substring from l to r inclusively (1 ≤ l ≤ r ≤ n), you should output
? l r
on a separate line. After this, all substrings of s[l..r] will be returned in random order, each substring exactly once. In every returned substring all characters will be randomly shuffled.
In the case, if you ask an incorrect query, ask more than 3 queries of the first type or there will be more than \left⌈ 0.777(n+1)^2 \right⌉ substrings returned in total, you will receive verdict Wrong answer.
To guess the string s, you should output
! s
on a separate line.
After printing each query, do not forget to flush the output. Otherwise, you will get Idleness limit exceeded. To flush the output, you can use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
If you received - (dash) as an answer to any query, you need to terminate your program with exit code 0 (for example, by calling exit(0)). This means that there was an error in the interaction protocol. If you don't terminate with exit code 0, you can receive any unsuccessful verdict.
Hack format
To hack a solution, use the following format:
The first line should contain one integer n (1 ≤ n ≤ 100) — the length of the string, and the following line should contain the string s.
Example
Input
4
a
aa
a
cb
b
c
c
Output
? 1 2
? 3 4
? 4 4
! aabc
Submitted Solution:
```
from sys import stdout
class String2:
def __init__(self, s, b=None):
if b is not None:
self.b = b
return
self.b = [0] * 26
for i in s:
self.b[ord(i) - ord('a')] += 1
def __add__(self, other):
b = self.b.copy()
for i in range(26):
b[i] += other.b[i]
return String2('', b)
def __sub__(self, other):
b = self.b.copy()
for i in range(26):
b[i] -= other.b[i]
return b
def __mul__(self, other):
ans = String2('', self.b)
for i in range(26):
ans.b[i] *= other
return ans
def req(l, r, k=0):
print('?', l, r)
v = [''.join(sorted(input())) for i in range((r - l + 1) * (r - l + 2) // 2)]
stdout.flush()
return v
def compute(v):
bukvi = [[0] * (n + 2) for _ in range(26)]
for el in v:
cur = len(el)
for e in el:
bukvi[ord(e) - ord('a')][cur] += 1
return bukvi
def compute2(bukvi):
bukvis = [set() for i in range(n + 2)]
for i in range(26):
prev = bukvi[i][1]
for j in range(1, n // 2 + n % 2 + 1):
while bukvi[i][j] != prev:
bukvis[j].add(chr(ord('a') + i))
prev += 1
return bukvis
def solve(va, va2):
for i in va2:
va.remove(i)
va.sort(key=len)
s = va[0]
for i in range(1, len(va)):
for j in range(26):
if va[i].count(chr(ord('a') + j)) != va[i - 1].count(chr(ord('a') + j)):
s += chr(ord('a') + j)
return s
def check(v, s, s2, f):
s3 = String2(v[0])
for i in range(1, len(v)):
s3 = s3 + String2(v[i])
le = len(v[0])
cur = String2('', String2('', f - String2(s)) - String2(s2)) * le
for i in range(le - 2):
cur = cur + (String2(s[i]) * (i + 1)) + (String2(s2[-i]) * (i + 1))
cur = cur + (String2(s[le - 2]) * (le - 1)) + (String2(s[le - 1]) * le)
e = cur - s3
for i in range(26):
if e[i]:
return chr(ord('a') + i)
def main():
va = req(1, n)
va2 = req(1, n // 2 + n % 2)
va3 = req(2, n // 2 + n % 2)
# bukvi2 = compute(va2)
# bukvi3 = compute(va3)
if n == 1:
print('!', va[0][0])
return
ma = [[] for i in range(n * 2)]
for i in va:
ma[len(i)].append(i)
a = String2(''.join(ma[1]))
s = solve(va2, va3)
s2 = ''
for i in range(2, n // 2 + 1):
s2 = check(ma[i], s, s2, a) + s2
se = String2('', a - String2(s)) - String2(s2)
for i in range(len(se)):
if se[i]:
s += chr(ord('a') + i)
break
print('!', s + s2)
n = int(input())
main()
```
No
| 103,044 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
N, K = map(int, input().split())
def f(k):
r = 0
for i in range(1, k):
r += i//2
return r
mx, mn = N+2, 0
idx = N//2
while mx-mn>1:
if f(idx) < K:
idx, mn = (idx+mx)//2, idx
continue
idx, mx = (idx+mn)//2, idx
#print(N, K)
#print(idx, f(idx), f(idx+1))
if idx+1 > N:
print(-1)
import sys
sys.exit()
rs = []
for i in range(idx):
rs.append(i+1)
rs.append(idx+1+2*(f(idx+1)-K))
#print(*rs)
rs2 = []
for i in range(N-(idx+1)):
rs2.append(5000+i)
rs = [10000*x for x in rs]
print(*(rs2 + rs))
```
| 103,045 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys
input = sys.stdin.readline
n,m=map(int,input().split())
x=(n-3)//2
MAX=x*(x+1)
if n%2==1:
MAX+=(n-1)//2
else:
MAX+=(n-1)//2*2
#print(MAX)
if m>MAX:
print(-1)
sys.exit()
score=0
x=1
while score<m:
x+=1
score+=(x-1)//2
#print(x,score)
LAST1=x
x+=(score-m)*2
ANS=list(range(1,LAST1))
ANS.append(x)
for i in range(n-len(ANS)):
ANS.append(10**9-i*25001-1)
print(*sorted(ANS))
```
| 103,046 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
def main():
n, m = map(int, input().split())
x = []
i = 1
while (n > 0 and m >= len(x) // 2):
m -= len(x) // 2
n -= 1
x.append(i)
i += 1
k = i - 1
if (m == 0 and n == 0):
print(*x)
return
elif (n == 0):
print(-1)
return
else:
x.append(2 * k - m * 2)
n -= 1
for i in range(n):
x.append(20000 * (i + 1) + 1)
print(*x)
main()
```
| 103,047 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
# from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var): sys.stdout.write(' '.join(map(str, var)) + '\n')
def out(var): sys.stdout.write(str(var) + '\n')
from decimal import Decimal
# from fractions import Fraction
# sys.setrecursionlimit(100000)
mod = int(1e9) + 7
INF=10**8
n,m=mdata()
if n==1 and m==0:
out(1)
exit()
l=[]
k=math.floor(((1+4*m)**0.5-1)/2)
if k>n:
out(-1)
exit()
l=list(range(1,2*k+3))
m-=k*(k+1)
while m:
l.append(l[-1]+l[max(0,l[-1]-2*m)])
m-=min(m,l[-1]//2)
if len(l)>n:
out(-1)
exit()
i=0
t=l[-1]
while len(l)<n:
l.append(INF+i*(t+1))
i+=1
outl(l)
```
| 103,048 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
# https://codeforces.com/problemset/problem/1305/E
def gen_sequence(size, balance):
result = []
for i in range(1, size + 1):
triple_count = (i - 1) >> 1
if triple_count <= balance:
result.append(i)
balance -= triple_count
else:
break
if len(result) == size and balance > 0:
return [-1]
if balance > 0:
result.append(2 * (result[-1] - balance) + 1)
delta = result[-1] + 1
while len(result) < size:
value = result[-1] + delta
if value % 2 == 0:
value += 1
result.append(value)
return result
if __name__ == '__main__':
size, balance = map(int, input().split())
for x in gen_sequence(size, balance):
print(x, end=' ')
print()
```
| 103,049 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
n, m = map(int, input().split())
out = []
currCount = 0
currVal = 0
while currVal < n:
nex = currVal // 2
if nex + currCount <= m:
currCount += nex
currVal += 1
out.append(currVal)
else:
break
need = m - currCount
if need > 0:
nex = currVal // 2
val = currVal + 1
val += 2 * (nex - need)
if nex > need:
out.append(val)
currCount += need
l = len(out)
if l > n or m != currCount:
assert(m > ((n-1) * (n-1))//4)
print(-1)
else:
assert(m <= ((n-1) * (n-1))//4)
lorg = max(out)
diff = lorg + 1
start = (lorg+diff)//diff
start += 2
start *= diff
start += 1
while l < n:
l += 1
out.append(start)
start += diff
assert(len(out) == n)
for i in range(n - 1):
assert(out[i] < out[i + 1])
assert(1 <= out[i])
assert(out[i + 1] <= 10 ** 9)
thing = set(out)
count = 0
for i in range(n):
for j in range(i + 1, n):
if out[i] + out[j] in thing:
count += 1
assert(count == m)
print(' '.join(map(str,out)))
```
| 103,050 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
n, m = map(int, input().split())
numList = [x+1 for x in range(n)]
backdoor = []
count = sum([(i-1) // 2 for i in range(1, n+1)])
if count < m: exit(print(-1))
while count > m:
lastpop = numList.pop()
count -= (lastpop - 1) // 2
if count >= m:
if len(backdoor) == 0: backdoor.append(10 ** 9)
else: backdoor.append(backdoor[-1] - 2 ** 16)
else:
gap = m - count
backdoor.append(2 * (lastpop - gap) - 1)
count += gap
while len(backdoor) > 0: numList.append(backdoor.pop())
print(' '.join([str(x) for x in numList]))
```
| 103,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Tags: constructive algorithms, greedy, implementation, math
Correct Solution:
```
import math
import sys
n,m=[int(s) for s in input().split()]
ans=[]
curr_balance=0
for i in range(1,n+1):
new_balance=math.floor((i-1)/2);
if curr_balance+new_balance > m:
break
curr_balance+=new_balance;
ans.append(i);
if curr_balance<m:
if len(ans)==n:
print(-1)
sys.exit()
remaining_balance = m-curr_balance
number_index = remaining_balance*2 - 1
if ans[-1-number_index]+ans[-1] <= 1000000000:
ans.append(ans[-1-number_index]+ans[-1])
num_to_add=ans[-1]+1
for i in range(len(ans)+1,n+1):
if ans[-1]+num_to_add <= 1000000000:
ans.append(ans[-1]+num_to_add)
else:
break;
if len(ans)<n:
print(-1)
sys.exit()
for i in ans:
print(i," ")
```
| 103,052 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,m=map(int,input().split())
x=(n-3)//2
MAX=x*(x+1)
if n%2==1:
MAX+=(n-1)//2
else:
MAX+=(n-1)//2*2
#print(MAX)
if m>MAX:
print(-1)
sys.exit()
score=0
x=1
while score<m:
x+=1
score+=(x-1)//2
#print(x,score)
LAST1=x
x+=(score-m)*2
ANS=list(range(1,LAST1))
ANS.append(x)
for i in range(n-len(ANS)):
ANS.append(10**9-i*5001-1)
print(*sorted(ANS))
```
Yes
| 103,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
from sys import stdin
from bisect import bisect_left as bl
from bisect import bisect_right as br
def input():
return stdin.readline()[:-1]
def intput():
return int(input())
def sinput():
return input().split()
def intsput():
return map(int, sinput())
class RangedList:
def __init__(self, start, stop, val=0):
self.shift = 0 - start
self.start = start
self.stop = stop
self.list = [val] * (stop - start)
def __setitem__(self, key, value):
self.list[key + self.shift] = value
def __getitem__(self, key):
return self.list[key + self.shift]
def __repr__(self):
return str(self.list)
def __iter__(self):
return iter(self.list)
def dprint(*args, **kwargs):
if debugging:
print(*args, **kwargs)
def sigma(x):
return x * (x + 1) // 2
def maxbalance(x):
if x % 2 == 0:
return sigma(x // 2 - 1) * 2
else:
x += 1
return sigma(x // 2 - 1) * 2 - x // 2 + 1
debugging = 1
# Code
n, balance = intsput()
if maxbalance(n) < balance:
print(-1)
exit()
else:
if n == 2:
print('1 2')
exit()
elif n == 1:
print('1')
exit()
largest = 2
dist = [1, 2]
k = 1
while len(dist) < n:
if balance:
x = dist[-1] + 1
can_create = (x - 1) // 2
if can_create <= balance:
dist.append(x)
balance -= can_create
largest = x
else:
dist.append(dist[- (balance * 2)] + dist[-1])
balance = 0
#used = set(dist)
else:
dist.append(10 ** 8 + 10 ** 4 * k + 1)
k += 1
print(*dist)
```
Yes
| 103,054 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
dic={}
count=0
for i in range(1,5001):
count+=(i-1)//2
dic[i]=count
n,m=map(int,input().split())
if(m>dic[n]):print(-1)
elif(m==0):
print(*[450000000+i for i in range(n)])
else:
ls=[]
flag=False
for i in range(1,5001):
if(dic[i]>m):
flag=True
break
else:
ls.append(i)
if(flag):m-=dic[i-1]
else:m-=dic[i]
if(m):ls.append(ls[-1]+ls[-2*m])
if(len(ls)<=n):
x=ls[-1]+ls[-2*m]
for i in range(n-len(ls)):
ls.append(10**7+x*i)
ls.sort()
print(*ls)
else:
print(-1)
```
Yes
| 103,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
n, m = map(int, input().split(' '))
limit = 0
for i in range(3, n + 1):
limit += (i - 1) // 2
if m > limit:
print(-1)
else:
a = [i for i in range(1, n + 1)]
count = limit
i = n
while count > m:
curr = (i - 1) // 2
to_del = min(curr, count - m)
if to_del == curr:
a[i - 1] = 1000000000 - (n - i) * 10000
else:
a[i - 1] = a[i - 2] + a[i - 1 - 2 * (curr - to_del)]
count -= to_del
i -= 1
print(' '.join(map(str, a)))
```
Yes
| 103,056 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
n,m = map(int,input().split())
a=[0]*5001
for i in range(1,5001):
a[i]=(i-1)//2
# print(a[:20])
for i in range(1,5001):
a[i]=a[i]+a[i-1]
# print(a[:20])
ans=[]
if m==0:
# ans.append(1)
aa=1
for i in range(n):
ans.append(aa)
aa=aa+4
print(*ans)
elif m>a[n]:
print(-1)
else:
if m in a:
ind = a.index(m)
for i in range(1,ind+1):
ans.append(i)
else:
for i in range(1,5000):
if a[i]<m<a[i+1]:
for j in range(1,i+1):
ans.append(j)
gap = m-a[i]
# print(gap,i)
ans.append(2*i-2*gap+1)
l=len(ans)
if l<n :
aa = ans[l-1]
x=aa+1
for k in range(n-l):
ans.append(aa+x)
aa = aa+x
print(len(ans))
print(*ans)
```
No
| 103,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
dic={}
count=0
for i in range(1,5001):
count+=(i-1)//2
dic[i]=count
n,m=map(int,input().split())
if(m>dic[n]):print(-1)
else:
ls=[]
for i in range(1,5000):
if(dic[i]>m):
break
else:
ls.append(i)
m-=dic[i-1]
while m and ls[-1]+ls[-2]<=10**9:
ls.append(ls[-1]+ls[-2])
m-=1
if(m):ls.append(10**9)
i=1
while m:
ls.append(10**9-i)
i+=1
m-=1
if(len(ls)<=n):
for i in range(n-len(ls)):
ls.append(10**9)
ls.sort()
print(*ls)
else:
print(-1)
```
No
| 103,058 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
from sys import stdin
from bisect import bisect_left as bl
from bisect import bisect_right as br
def input():
return stdin.readline()[:-1]
def intput():
return int(input())
def sinput():
return input().split()
def intsput():
return map(int, sinput())
class RangedList:
def __init__(self, start, stop, val=0):
self.shift = 0 - start
self.start = start
self.stop = stop
self.list = [val] * (stop - start)
def __setitem__(self, key, value):
self.list[key + self.shift] = value
def __getitem__(self, key):
return self.list[key + self.shift]
def __repr__(self):
return str(self.list)
def __iter__(self):
return iter(self.list)
def dprint(*args, **kwargs):
if debugging:
print(*args, **kwargs)
def sigma(x):
return x * (x + 1) // 2
def maxbalance(x):
if x % 2 == 0:
return sigma(x // 2 - 1) * 2
else:
x += 1
return sigma(x // 2 - 1) * 2 - x // 2 + 1
debugging = 1
# Code
n, balance = intsput()
if maxbalance(n) < balance:
print(-1)
exit()
else:
if n <= 2:
print('1 2')
exit()
largest = 2
dist = [1, 2]
k = 1
while len(dist) < n:
if balance:
x = dist[-1] + 1
can_create = (x - 1) // 2
if can_create <= balance:
dist.append(x)
balance -= can_create
largest = x
else:
dist.append(dist[- (balance * 2)] + dist[-1])
balance = 0
#used = set(dist)
else:
dist.append(10 ** 8 + 10 ** 4 * k + 1)
k += 1
print(*dist)
```
No
| 103,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is the coordinator of the next Mathforces round written by the "Proof by AC" team. All the preparation has been done, and he is discussing with the team about the score distribution for the round.
The round consists of n problems, numbered from 1 to n. The problems are ordered in increasing order of difficulty, no two problems have the same difficulty. A score distribution for the round can be denoted by an array a_1, a_2, ..., a_n, where a_i is the score of i-th problem.
Kuroni thinks that the score distribution should satisfy the following requirements:
* The score of each problem should be a positive integer not exceeding 10^9.
* A harder problem should grant a strictly higher score than an easier problem. In other words, 1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9.
* The balance of the score distribution, defined as the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and a_i + a_j = a_k, should be exactly m.
Help the team find a score distribution that satisfies Kuroni's requirement. In case such a score distribution does not exist, output -1.
Input
The first and single line contains two integers n and m (1 ≤ n ≤ 5000, 0 ≤ m ≤ 10^9) — the number of problems and the required balance.
Output
If there is no solution, print a single integer -1.
Otherwise, print a line containing n integers a_1, a_2, ..., a_n, representing a score distribution that satisfies all the requirements. If there are multiple answers, print any of them.
Examples
Input
5 3
Output
4 5 9 13 18
Input
8 0
Output
10 11 12 13 14 15 16 17
Input
4 10
Output
-1
Note
In the first example, there are 3 triples (i, j, k) that contribute to the balance of the score distribution.
* (1, 2, 3)
* (1, 3, 4)
* (2, 4, 5)
Submitted Solution:
```
def Fib(n):
f = [1,2]
for i in range(3,n+1):
x = f[-1]+f[-2]
f.append(x)
return f
fib = Fib(10001)
# print(fib)
def f():
n, m = [int(s) for s in input().split()]
if m >=n-1:
print(-1)
return
ans = fib[:m+2]
i = m
for j in range(n-(m+2)):
ans.append(fib[i])
i += 2
print(' '.join(str(num) for num in ans))
f()
```
No
| 103,060 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya very much likes playing with little Masha. Recently he has received a game called "Zero-One" as a gift from his mother. Petya immediately offered Masha to play the game with him.
Before the very beginning of the game several cards are lain out on a table in one line from the left to the right. Each card contains a digit: 0 or 1. Players move in turns and Masha moves first. During each move a player should remove a card from the table and shift all other cards so as to close the gap left by the removed card. For example, if before somebody's move the cards on the table formed a sequence 01010101, then after the fourth card is removed (the cards are numbered starting from 1), the sequence will look like that: 0100101.
The game ends when exactly two cards are left on the table. The digits on these cards determine the number in binary notation: the most significant bit is located to the left. Masha's aim is to minimize the number and Petya's aim is to maximize it.
An unpleasant accident occurred before the game started. The kids spilled juice on some of the cards and the digits on the cards got blurred. Each one of the spoiled cards could have either 0 or 1 written on it. Consider all possible variants of initial arrangement of the digits (before the juice spilling). For each variant, let's find which two cards are left by the end of the game, assuming that both Petya and Masha play optimally. An ordered pair of digits written on those two cards is called an outcome. Your task is to find the set of outcomes for all variants of initial digits arrangement.
Input
The first line contains a sequence of characters each of which can either be a "0", a "1" or a "?". This sequence determines the initial arrangement of cards on the table from the left to the right. The characters "?" mean that the given card was spoiled before the game. The sequence's length ranges from 2 to 105, inclusive.
Output
Print the set of outcomes for all possible initial digits arrangements. Print each possible outcome on a single line. Each outcome should be represented by two characters: the digits written on the cards that were left by the end of the game. The outcomes should be sorted lexicographically in ascending order (see the first sample).
Examples
Input
????
Output
00
01
10
11
Input
1010
Output
10
Input
1?1
Output
01
11
Note
In the first sample all 16 variants of numbers arrangement are possible. For the variant 0000 the outcome is 00. For the variant 1111 the outcome is 11. For the variant 0011 the outcome is 01. For the variant 1100 the outcome is 10. Regardless of outcomes for all other variants the set which we are looking for will contain all 4 possible outcomes.
In the third sample only 2 variants of numbers arrangement are possible: 111 and 101. For the variant 111 the outcome is 11. For the variant 101 the outcome is 01, because on the first turn Masha can remove the first card from the left after which the game will end.
Tags: constructive algorithms, games, greedy
Correct Solution:
```
def evaluate(a):
c1 = a.count('1')
c0 = a.count('0')
n = len(a)
A = (n - 1) // 2
B = (n - 2) // 2
if c1 <= A:
return '00'
if c0 <= B:
return '11'
p1 = a.rfind('1')
p0 = a.rfind('0')
if p0 < p1:
return '01'
else:
return '10'
a = input()
x = []
x.append(evaluate(a.replace('?', '0')))
x.append(evaluate(a.replace('?', '1')))
n = len(a)
c1 = a.count('1')
c0 = a.count('0')
A = (n - 1) // 2
B = (n - 2) // 2
x.append(evaluate(a.replace('?', '0', B + 1 - c0).replace('?', '1')))
x.append(evaluate(a.replace('?', '1', A + 1 - c1).replace('?', '0')))
for ans in sorted(list(set(x))):
print(ans)
```
| 103,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Omkar has just come across a duck! The duck is walking on a grid with n rows and n columns (2 ≤ n ≤ 25) so that the grid contains a total of n^2 cells. Let's denote by (x, y) the cell in the x-th row from the top and the y-th column from the left. Right now, the duck is at the cell (1, 1) (the cell in the top left corner) and would like to reach the cell (n, n) (the cell in the bottom right corner) by moving either down 1 cell or to the right 1 cell each second.
Since Omkar thinks ducks are fun, he wants to play a game with you based on the movement of the duck. First, for each cell (x, y) in the grid, you will tell Omkar a nonnegative integer a_{x,y} not exceeding 10^{16}, and Omkar will then put a_{x,y} uninteresting problems in the cell (x, y). After that, the duck will start their journey from (1, 1) to (n, n). For each cell (x, y) that the duck crosses during their journey (including the cells (1, 1) and (n, n)), the duck will eat the a_{x,y} uninteresting problems in that cell. Once the duck has completed their journey, Omkar will measure their mass to determine the total number k of uninteresting problems that the duck ate on their journey, and then tell you k.
Your challenge, given k, is to exactly reproduce the duck's path, i. e. to tell Omkar precisely which cells the duck crossed on their journey. To be sure of your mastery of this game, Omkar will have the duck complete q different journeys (1 ≤ q ≤ 10^3). Note that all journeys are independent: at the beginning of each journey, the cell (x, y) will still contain a_{x,y} uninteresting tasks.
Interaction
The interaction will begin with a line containing a single integer n (2 ≤ n ≤ 25), the amount of rows and columns in the grid. Read it.
Your program should then print n lines. The x-th line should contain n integers a_{x,1}, a_{x,2}, ..., a_{x,n} satisfying 0 ≤ a_{x,y} ≤ 10^{16}, where a_{x,y} is the amount of uninteresting problems Omkar should place in the cell (x, y).
After that, you will first receive a single integer q, the amount of journeys that the duck will take. q queries will follow; each query will consist of a single line containing an integer k, the amount of uninteresting problems that the duck ate on that journey. After each query, given that you have determined that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order (it should always be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n)), you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j.
Bear in mind that if the sum on your path is k, but your path is different from the actual hidden path, then your solution is still wrong!
After printing each line do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, first output a line containing n and another line containing q. It must be true that 2 ≤ n ≤ 25 and 1 ≤ q ≤ 1000. Then, output the q journeys taken by the duck in the same format as described above: for each journey, given that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order, you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j. It must be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n). Additionally, for each j such that 2 ≤ j ≤ 2n - 1, it must be true that 1 ≤ x_j, y_j ≤ n and either (x_j, y_j) = (x_{j - 1} + 1, y_{j - 1}) or (x_j, y_j) = (x_{j - 1}, y_{j - 1} + 1).
Example
Input
4
3
23
26
27
Output
1 2 3 6
4 6 2 10
9 0 7 3
2 8 8 2
1 1
1 2
1 3
2 3
2 4
3 4
4 4
1 1
2 1
3 1
3 2
3 3
3 4
4 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4
Note
The duck's three journeys are illustrated below.
1 + 2 + 3 + 2 + 10 + 3 + 2 = 23
<image>
1 + 4 + 9 + 0 + 7 + 3 + 2 = 26
<image>
1 + 2 + 3 + 6 + 10 + 3 + 2 = 27
<image>
Tags: bitmasks, constructive algorithms, interactive, math
Correct Solution:
```
import sys
def pow(i,j):
s = i+j
if i % 2 ==0:
return 0
return 2**s
def nxt():
s = input()
while not s:
s = input()
return int(s)
def main():
(n,) = map(int, input().split(' '))
a = []
for i in range(n):
b = []
for j in range(n):
z = pow(i,j)
b.append(z)
print(z, end=" ")
a.append(b)
print()
sys.stdout.flush()
for _ in range(nxt()):
k = nxt()
i = 0
j = 0
print(1, 1)
while i < n - 1 or j < n - 1:
if i == n - 1:
j += 1
elif j == n - 1:
i += 1
else:
if a[i + 1][j]>0:
if a[i + 1][j] & k == a[i + 1][j]:
i += 1
else:
j += 1
else:
if a[i ][j+1] & k == a[i ][j+1]:
j += 1
else:
i += 1
print(i + 1, j + 1)
print()
sys.stdout.flush()
# for _ in range(int(input())):
main()
```
| 103,062 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Omkar has just come across a duck! The duck is walking on a grid with n rows and n columns (2 ≤ n ≤ 25) so that the grid contains a total of n^2 cells. Let's denote by (x, y) the cell in the x-th row from the top and the y-th column from the left. Right now, the duck is at the cell (1, 1) (the cell in the top left corner) and would like to reach the cell (n, n) (the cell in the bottom right corner) by moving either down 1 cell or to the right 1 cell each second.
Since Omkar thinks ducks are fun, he wants to play a game with you based on the movement of the duck. First, for each cell (x, y) in the grid, you will tell Omkar a nonnegative integer a_{x,y} not exceeding 10^{16}, and Omkar will then put a_{x,y} uninteresting problems in the cell (x, y). After that, the duck will start their journey from (1, 1) to (n, n). For each cell (x, y) that the duck crosses during their journey (including the cells (1, 1) and (n, n)), the duck will eat the a_{x,y} uninteresting problems in that cell. Once the duck has completed their journey, Omkar will measure their mass to determine the total number k of uninteresting problems that the duck ate on their journey, and then tell you k.
Your challenge, given k, is to exactly reproduce the duck's path, i. e. to tell Omkar precisely which cells the duck crossed on their journey. To be sure of your mastery of this game, Omkar will have the duck complete q different journeys (1 ≤ q ≤ 10^3). Note that all journeys are independent: at the beginning of each journey, the cell (x, y) will still contain a_{x,y} uninteresting tasks.
Interaction
The interaction will begin with a line containing a single integer n (2 ≤ n ≤ 25), the amount of rows and columns in the grid. Read it.
Your program should then print n lines. The x-th line should contain n integers a_{x,1}, a_{x,2}, ..., a_{x,n} satisfying 0 ≤ a_{x,y} ≤ 10^{16}, where a_{x,y} is the amount of uninteresting problems Omkar should place in the cell (x, y).
After that, you will first receive a single integer q, the amount of journeys that the duck will take. q queries will follow; each query will consist of a single line containing an integer k, the amount of uninteresting problems that the duck ate on that journey. After each query, given that you have determined that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order (it should always be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n)), you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j.
Bear in mind that if the sum on your path is k, but your path is different from the actual hidden path, then your solution is still wrong!
After printing each line do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, first output a line containing n and another line containing q. It must be true that 2 ≤ n ≤ 25 and 1 ≤ q ≤ 1000. Then, output the q journeys taken by the duck in the same format as described above: for each journey, given that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order, you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j. It must be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n). Additionally, for each j such that 2 ≤ j ≤ 2n - 1, it must be true that 1 ≤ x_j, y_j ≤ n and either (x_j, y_j) = (x_{j - 1} + 1, y_{j - 1}) or (x_j, y_j) = (x_{j - 1}, y_{j - 1} + 1).
Example
Input
4
3
23
26
27
Output
1 2 3 6
4 6 2 10
9 0 7 3
2 8 8 2
1 1
1 2
1 3
2 3
2 4
3 4
4 4
1 1
2 1
3 1
3 2
3 3
3 4
4 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4
Note
The duck's three journeys are illustrated below.
1 + 2 + 3 + 2 + 10 + 3 + 2 = 23
<image>
1 + 4 + 9 + 0 + 7 + 3 + 2 = 26
<image>
1 + 2 + 3 + 6 + 10 + 3 + 2 = 27
<image>
Tags: bitmasks, constructive algorithms, interactive, math
Correct Solution:
```
n = int(input().split()[0])
arr = [[0]*n for x in range(n)]
##for x in range(n):
## for y in range(n):
## if x % 2 == 0 and y <= n-x:
## arr[x][y] = 2**(x+y-1)
## if y % 2 == 0 and x <= n-y:
## arr[x][y] = 2**(x+y-1)
for d in range(1,2*n-2):
if d < n-1:
curr = True
for x in range(d+1):
if curr:
arr[x][d-x] = 2**(d-1)
curr = not curr
else:
curr = True
for x in range(n-1,d-n,-1):
if curr:
arr[d-x][x] = 2**(d-1)
curr = not curr
for x in range(n):
print(*arr[x],flush=True)
q = int(input().split()[0])
for query in range(q):
tot = int(input().split()[0])
ans = []
ans.append([0,0])
currx = 0
curry = 0
for i in range(2*n-2):
if currx == n-1:
curry += 1
while curry < n:
ans.append([currx,curry])
curry += 1
break
if curry == n-1:
currx += 1
while currx < n:
ans.append([currx,curry])
currx += 1
break
if tot & (2**i) != 2**i:
if arr[currx+1][curry] == 0:
currx += 1
else:
curry += 1
else:
if arr[currx+1][curry] == 0:
curry += 1
else:
currx += 1
ans.append([currx,curry])
for i in range(2*n-1):
print(str(ans[i][0]+1) + " " + str(ans[i][1] + 1),flush=True)
```
| 103,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Omkar has just come across a duck! The duck is walking on a grid with n rows and n columns (2 ≤ n ≤ 25) so that the grid contains a total of n^2 cells. Let's denote by (x, y) the cell in the x-th row from the top and the y-th column from the left. Right now, the duck is at the cell (1, 1) (the cell in the top left corner) and would like to reach the cell (n, n) (the cell in the bottom right corner) by moving either down 1 cell or to the right 1 cell each second.
Since Omkar thinks ducks are fun, he wants to play a game with you based on the movement of the duck. First, for each cell (x, y) in the grid, you will tell Omkar a nonnegative integer a_{x,y} not exceeding 10^{16}, and Omkar will then put a_{x,y} uninteresting problems in the cell (x, y). After that, the duck will start their journey from (1, 1) to (n, n). For each cell (x, y) that the duck crosses during their journey (including the cells (1, 1) and (n, n)), the duck will eat the a_{x,y} uninteresting problems in that cell. Once the duck has completed their journey, Omkar will measure their mass to determine the total number k of uninteresting problems that the duck ate on their journey, and then tell you k.
Your challenge, given k, is to exactly reproduce the duck's path, i. e. to tell Omkar precisely which cells the duck crossed on their journey. To be sure of your mastery of this game, Omkar will have the duck complete q different journeys (1 ≤ q ≤ 10^3). Note that all journeys are independent: at the beginning of each journey, the cell (x, y) will still contain a_{x,y} uninteresting tasks.
Interaction
The interaction will begin with a line containing a single integer n (2 ≤ n ≤ 25), the amount of rows and columns in the grid. Read it.
Your program should then print n lines. The x-th line should contain n integers a_{x,1}, a_{x,2}, ..., a_{x,n} satisfying 0 ≤ a_{x,y} ≤ 10^{16}, where a_{x,y} is the amount of uninteresting problems Omkar should place in the cell (x, y).
After that, you will first receive a single integer q, the amount of journeys that the duck will take. q queries will follow; each query will consist of a single line containing an integer k, the amount of uninteresting problems that the duck ate on that journey. After each query, given that you have determined that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order (it should always be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n)), you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j.
Bear in mind that if the sum on your path is k, but your path is different from the actual hidden path, then your solution is still wrong!
After printing each line do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, first output a line containing n and another line containing q. It must be true that 2 ≤ n ≤ 25 and 1 ≤ q ≤ 1000. Then, output the q journeys taken by the duck in the same format as described above: for each journey, given that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order, you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j. It must be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n). Additionally, for each j such that 2 ≤ j ≤ 2n - 1, it must be true that 1 ≤ x_j, y_j ≤ n and either (x_j, y_j) = (x_{j - 1} + 1, y_{j - 1}) or (x_j, y_j) = (x_{j - 1}, y_{j - 1} + 1).
Example
Input
4
3
23
26
27
Output
1 2 3 6
4 6 2 10
9 0 7 3
2 8 8 2
1 1
1 2
1 3
2 3
2 4
3 4
4 4
1 1
2 1
3 1
3 2
3 3
3 4
4 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4
Note
The duck's three journeys are illustrated below.
1 + 2 + 3 + 2 + 10 + 3 + 2 = 23
<image>
1 + 4 + 9 + 0 + 7 + 3 + 2 = 26
<image>
1 + 2 + 3 + 6 + 10 + 3 + 2 = 27
<image>
Tags: bitmasks, constructive algorithms, interactive, math
Correct Solution:
```
def read():
s = ''
while not s.strip().isdigit():
s = input()
return int(s)
def tile(i, j):
return 2 ** (i + j) if i % 2 else 0
n = read()
for i in range(n):
for x in [tile(i, j) for j in range(n)]:
print(x, end = ' ')
print('', flush = True)
q = read()
for qx in range(q):
k = read()
i = n - 1; j = n - 1
ans = []
while i > 0 or j > 0:
ans.append((i + 1, j + 1))
k -= tile(i, j)
if i == 0:
j -= 1
elif j == 0:
i -= 1
else:
if (i % 2 and k & tile(i, j - 1)) or (i % 2 == 0 and (k & tile(i - 1, j)) == 0):
j -= 1
else:
i -= 1
ans.append((1, 1))
for x in reversed(ans):
print(f'{x[0]} {x[1]}')
print('', flush = True)
```
| 103,064 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Omkar has just come across a duck! The duck is walking on a grid with n rows and n columns (2 ≤ n ≤ 25) so that the grid contains a total of n^2 cells. Let's denote by (x, y) the cell in the x-th row from the top and the y-th column from the left. Right now, the duck is at the cell (1, 1) (the cell in the top left corner) and would like to reach the cell (n, n) (the cell in the bottom right corner) by moving either down 1 cell or to the right 1 cell each second.
Since Omkar thinks ducks are fun, he wants to play a game with you based on the movement of the duck. First, for each cell (x, y) in the grid, you will tell Omkar a nonnegative integer a_{x,y} not exceeding 10^{16}, and Omkar will then put a_{x,y} uninteresting problems in the cell (x, y). After that, the duck will start their journey from (1, 1) to (n, n). For each cell (x, y) that the duck crosses during their journey (including the cells (1, 1) and (n, n)), the duck will eat the a_{x,y} uninteresting problems in that cell. Once the duck has completed their journey, Omkar will measure their mass to determine the total number k of uninteresting problems that the duck ate on their journey, and then tell you k.
Your challenge, given k, is to exactly reproduce the duck's path, i. e. to tell Omkar precisely which cells the duck crossed on their journey. To be sure of your mastery of this game, Omkar will have the duck complete q different journeys (1 ≤ q ≤ 10^3). Note that all journeys are independent: at the beginning of each journey, the cell (x, y) will still contain a_{x,y} uninteresting tasks.
Interaction
The interaction will begin with a line containing a single integer n (2 ≤ n ≤ 25), the amount of rows and columns in the grid. Read it.
Your program should then print n lines. The x-th line should contain n integers a_{x,1}, a_{x,2}, ..., a_{x,n} satisfying 0 ≤ a_{x,y} ≤ 10^{16}, where a_{x,y} is the amount of uninteresting problems Omkar should place in the cell (x, y).
After that, you will first receive a single integer q, the amount of journeys that the duck will take. q queries will follow; each query will consist of a single line containing an integer k, the amount of uninteresting problems that the duck ate on that journey. After each query, given that you have determined that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order (it should always be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n)), you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j.
Bear in mind that if the sum on your path is k, but your path is different from the actual hidden path, then your solution is still wrong!
After printing each line do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, first output a line containing n and another line containing q. It must be true that 2 ≤ n ≤ 25 and 1 ≤ q ≤ 1000. Then, output the q journeys taken by the duck in the same format as described above: for each journey, given that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order, you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j. It must be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n). Additionally, for each j such that 2 ≤ j ≤ 2n - 1, it must be true that 1 ≤ x_j, y_j ≤ n and either (x_j, y_j) = (x_{j - 1} + 1, y_{j - 1}) or (x_j, y_j) = (x_{j - 1}, y_{j - 1} + 1).
Example
Input
4
3
23
26
27
Output
1 2 3 6
4 6 2 10
9 0 7 3
2 8 8 2
1 1
1 2
1 3
2 3
2 4
3 4
4 4
1 1
2 1
3 1
3 2
3 3
3 4
4 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4
Note
The duck's three journeys are illustrated below.
1 + 2 + 3 + 2 + 10 + 3 + 2 = 23
<image>
1 + 4 + 9 + 0 + 7 + 3 + 2 = 26
<image>
1 + 2 + 3 + 6 + 10 + 3 + 2 = 27
<image>
Submitted Solution:
```
def main():
n = int(input())
for i in range(n):
print([x*i + x for x in range(n)])
# paths = set()
# for i in range(n):
# solve()
def solve():
n = int(input())
main()
```
No
| 103,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Omkar has just come across a duck! The duck is walking on a grid with n rows and n columns (2 ≤ n ≤ 25) so that the grid contains a total of n^2 cells. Let's denote by (x, y) the cell in the x-th row from the top and the y-th column from the left. Right now, the duck is at the cell (1, 1) (the cell in the top left corner) and would like to reach the cell (n, n) (the cell in the bottom right corner) by moving either down 1 cell or to the right 1 cell each second.
Since Omkar thinks ducks are fun, he wants to play a game with you based on the movement of the duck. First, for each cell (x, y) in the grid, you will tell Omkar a nonnegative integer a_{x,y} not exceeding 10^{16}, and Omkar will then put a_{x,y} uninteresting problems in the cell (x, y). After that, the duck will start their journey from (1, 1) to (n, n). For each cell (x, y) that the duck crosses during their journey (including the cells (1, 1) and (n, n)), the duck will eat the a_{x,y} uninteresting problems in that cell. Once the duck has completed their journey, Omkar will measure their mass to determine the total number k of uninteresting problems that the duck ate on their journey, and then tell you k.
Your challenge, given k, is to exactly reproduce the duck's path, i. e. to tell Omkar precisely which cells the duck crossed on their journey. To be sure of your mastery of this game, Omkar will have the duck complete q different journeys (1 ≤ q ≤ 10^3). Note that all journeys are independent: at the beginning of each journey, the cell (x, y) will still contain a_{x,y} uninteresting tasks.
Interaction
The interaction will begin with a line containing a single integer n (2 ≤ n ≤ 25), the amount of rows and columns in the grid. Read it.
Your program should then print n lines. The x-th line should contain n integers a_{x,1}, a_{x,2}, ..., a_{x,n} satisfying 0 ≤ a_{x,y} ≤ 10^{16}, where a_{x,y} is the amount of uninteresting problems Omkar should place in the cell (x, y).
After that, you will first receive a single integer q, the amount of journeys that the duck will take. q queries will follow; each query will consist of a single line containing an integer k, the amount of uninteresting problems that the duck ate on that journey. After each query, given that you have determined that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order (it should always be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n)), you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j.
Bear in mind that if the sum on your path is k, but your path is different from the actual hidden path, then your solution is still wrong!
After printing each line do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, first output a line containing n and another line containing q. It must be true that 2 ≤ n ≤ 25 and 1 ≤ q ≤ 1000. Then, output the q journeys taken by the duck in the same format as described above: for each journey, given that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order, you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j. It must be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n). Additionally, for each j such that 2 ≤ j ≤ 2n - 1, it must be true that 1 ≤ x_j, y_j ≤ n and either (x_j, y_j) = (x_{j - 1} + 1, y_{j - 1}) or (x_j, y_j) = (x_{j - 1}, y_{j - 1} + 1).
Example
Input
4
3
23
26
27
Output
1 2 3 6
4 6 2 10
9 0 7 3
2 8 8 2
1 1
1 2
1 3
2 3
2 4
3 4
4 4
1 1
2 1
3 1
3 2
3 3
3 4
4 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4
Note
The duck's three journeys are illustrated below.
1 + 2 + 3 + 2 + 10 + 3 + 2 = 23
<image>
1 + 4 + 9 + 0 + 7 + 3 + 2 = 26
<image>
1 + 2 + 3 + 6 + 10 + 3 + 2 = 27
<image>
Submitted Solution:
```
n = int(input())
a = [2**(i) for i in range(n**2)]
b = []
for j in range(n):
c = []
for k in range(n):
c.append(a[j*n+k])
b.append(c)
for row in b:
print(" ".join(str(num) for num in row), flush=True)
for i in range(int(input())):
q = int(input())
x = 0
y = 0
while (x < n-1 or y < n-1):
print(f"{y+1} {x+1}", flush=True)
#Can we go right?
if x == n-1:
y += 1
elif b[y][x+1]&q:
x += 1
else:
y += 1
print(f"{n} {n}",flush=True)
```
No
| 103,066 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Omkar has just come across a duck! The duck is walking on a grid with n rows and n columns (2 ≤ n ≤ 25) so that the grid contains a total of n^2 cells. Let's denote by (x, y) the cell in the x-th row from the top and the y-th column from the left. Right now, the duck is at the cell (1, 1) (the cell in the top left corner) and would like to reach the cell (n, n) (the cell in the bottom right corner) by moving either down 1 cell or to the right 1 cell each second.
Since Omkar thinks ducks are fun, he wants to play a game with you based on the movement of the duck. First, for each cell (x, y) in the grid, you will tell Omkar a nonnegative integer a_{x,y} not exceeding 10^{16}, and Omkar will then put a_{x,y} uninteresting problems in the cell (x, y). After that, the duck will start their journey from (1, 1) to (n, n). For each cell (x, y) that the duck crosses during their journey (including the cells (1, 1) and (n, n)), the duck will eat the a_{x,y} uninteresting problems in that cell. Once the duck has completed their journey, Omkar will measure their mass to determine the total number k of uninteresting problems that the duck ate on their journey, and then tell you k.
Your challenge, given k, is to exactly reproduce the duck's path, i. e. to tell Omkar precisely which cells the duck crossed on their journey. To be sure of your mastery of this game, Omkar will have the duck complete q different journeys (1 ≤ q ≤ 10^3). Note that all journeys are independent: at the beginning of each journey, the cell (x, y) will still contain a_{x,y} uninteresting tasks.
Interaction
The interaction will begin with a line containing a single integer n (2 ≤ n ≤ 25), the amount of rows and columns in the grid. Read it.
Your program should then print n lines. The x-th line should contain n integers a_{x,1}, a_{x,2}, ..., a_{x,n} satisfying 0 ≤ a_{x,y} ≤ 10^{16}, where a_{x,y} is the amount of uninteresting problems Omkar should place in the cell (x, y).
After that, you will first receive a single integer q, the amount of journeys that the duck will take. q queries will follow; each query will consist of a single line containing an integer k, the amount of uninteresting problems that the duck ate on that journey. After each query, given that you have determined that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order (it should always be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n)), you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j.
Bear in mind that if the sum on your path is k, but your path is different from the actual hidden path, then your solution is still wrong!
After printing each line do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, first output a line containing n and another line containing q. It must be true that 2 ≤ n ≤ 25 and 1 ≤ q ≤ 1000. Then, output the q journeys taken by the duck in the same format as described above: for each journey, given that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order, you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j. It must be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n). Additionally, for each j such that 2 ≤ j ≤ 2n - 1, it must be true that 1 ≤ x_j, y_j ≤ n and either (x_j, y_j) = (x_{j - 1} + 1, y_{j - 1}) or (x_j, y_j) = (x_{j - 1}, y_{j - 1} + 1).
Example
Input
4
3
23
26
27
Output
1 2 3 6
4 6 2 10
9 0 7 3
2 8 8 2
1 1
1 2
1 3
2 3
2 4
3 4
4 4
1 1
2 1
3 1
3 2
3 3
3 4
4 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4
Note
The duck's three journeys are illustrated below.
1 + 2 + 3 + 2 + 10 + 3 + 2 = 23
<image>
1 + 4 + 9 + 0 + 7 + 3 + 2 = 26
<image>
1 + 2 + 3 + 6 + 10 + 3 + 2 = 27
<image>
Submitted Solution:
```
size = int(input())
#size =
dp = [[1 for i in range(size)] for j in range(size)]
for i in range(1,size):
for j in range(1,size):
dp[i][j] = dp[i-1][j] + dp[i][j-1]
#dp[j][-1] = [0]
#print(dp)
#s()
#for case in range(T):
maze = [[0 for i in range(size)] for j in range(size)]
#maze[1][1:] = [2**i
d = {i:{j:[] for j in range(-1,size)}for i in range(-1,size)}
for j in range(-1,size):
d[-1][j] = [0]
d[j][-1] = [0]
m = 0
for i in range(1,size):
for j in range(size-2,-1,-1):
#print(i,j)
last = dp[size-j-2][-1]
lastsmall = 0
al = (sum(maze[i][:]) + sum([maze[x][j] for x in range(size)]))
for k in range(size-2,size-1-i,-1):
lastsmall += dp[size-j-2][k]
#print(i,j,last,lastsmall,al)#,maze[i][:],maze[:][j])
maze[i][j] = last + lastsmall - al
#print(maze)
#d[i][j]+=[x+maze[i][j] for x in d[i-1][j]]
#d[i][j]+=[x+maze[i][j] for x in d[i][j-1]]
for i in range(size):
print(" ".join([str(x) for x in maze[i]]),flush = True)
#stdout.flush()
qs = int(input())
for q in range(qs):
ss = int(input())
si = 0
sj = 0
tot = 0
while si != size - 1 or sj != size - 1:
print(si+1,sj+1,flush = True)
tot += maze[si][sj]
if si == size-1:
sj += 1
elif sj == size - 1:
si += 1
elif tot + maze[si+1][sj] > ss:
sj += 1
else:
si += 1
print(si+1,sj+1,flush = True)
#print("e")
#print(size[i][j])
```
No
| 103,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Omkar has just come across a duck! The duck is walking on a grid with n rows and n columns (2 ≤ n ≤ 25) so that the grid contains a total of n^2 cells. Let's denote by (x, y) the cell in the x-th row from the top and the y-th column from the left. Right now, the duck is at the cell (1, 1) (the cell in the top left corner) and would like to reach the cell (n, n) (the cell in the bottom right corner) by moving either down 1 cell or to the right 1 cell each second.
Since Omkar thinks ducks are fun, he wants to play a game with you based on the movement of the duck. First, for each cell (x, y) in the grid, you will tell Omkar a nonnegative integer a_{x,y} not exceeding 10^{16}, and Omkar will then put a_{x,y} uninteresting problems in the cell (x, y). After that, the duck will start their journey from (1, 1) to (n, n). For each cell (x, y) that the duck crosses during their journey (including the cells (1, 1) and (n, n)), the duck will eat the a_{x,y} uninteresting problems in that cell. Once the duck has completed their journey, Omkar will measure their mass to determine the total number k of uninteresting problems that the duck ate on their journey, and then tell you k.
Your challenge, given k, is to exactly reproduce the duck's path, i. e. to tell Omkar precisely which cells the duck crossed on their journey. To be sure of your mastery of this game, Omkar will have the duck complete q different journeys (1 ≤ q ≤ 10^3). Note that all journeys are independent: at the beginning of each journey, the cell (x, y) will still contain a_{x,y} uninteresting tasks.
Interaction
The interaction will begin with a line containing a single integer n (2 ≤ n ≤ 25), the amount of rows and columns in the grid. Read it.
Your program should then print n lines. The x-th line should contain n integers a_{x,1}, a_{x,2}, ..., a_{x,n} satisfying 0 ≤ a_{x,y} ≤ 10^{16}, where a_{x,y} is the amount of uninteresting problems Omkar should place in the cell (x, y).
After that, you will first receive a single integer q, the amount of journeys that the duck will take. q queries will follow; each query will consist of a single line containing an integer k, the amount of uninteresting problems that the duck ate on that journey. After each query, given that you have determined that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order (it should always be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n)), you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j.
Bear in mind that if the sum on your path is k, but your path is different from the actual hidden path, then your solution is still wrong!
After printing each line do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, first output a line containing n and another line containing q. It must be true that 2 ≤ n ≤ 25 and 1 ≤ q ≤ 1000. Then, output the q journeys taken by the duck in the same format as described above: for each journey, given that the duck visited the cells (x_1, y_1), (x_2, y_2), ..., (x_{2n - 1}, y_{2n - 1}) in that order, you should output 2n - 1 lines so that the j-th line contains the two integers x_j, y_j. It must be true that (x_1, y_1) = (1, 1) and (x_{2n - 1}, y_{2n - 1}) = (n, n). Additionally, for each j such that 2 ≤ j ≤ 2n - 1, it must be true that 1 ≤ x_j, y_j ≤ n and either (x_j, y_j) = (x_{j - 1} + 1, y_{j - 1}) or (x_j, y_j) = (x_{j - 1}, y_{j - 1} + 1).
Example
Input
4
3
23
26
27
Output
1 2 3 6
4 6 2 10
9 0 7 3
2 8 8 2
1 1
1 2
1 3
2 3
2 4
3 4
4 4
1 1
2 1
3 1
3 2
3 3
3 4
4 4
1 1
1 2
1 3
1 4
2 4
3 4
4 4
Note
The duck's three journeys are illustrated below.
1 + 2 + 3 + 2 + 10 + 3 + 2 = 23
<image>
1 + 4 + 9 + 0 + 7 + 3 + 2 = 26
<image>
1 + 2 + 3 + 6 + 10 + 3 + 2 = 27
<image>
Submitted Solution:
```
import time,math as mt,bisect,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IO(): # to take string input
return stdin.readline()
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def readTree(): # to read tree
v=int(input())
adj=[set() for i in range(v+1)]
for i in range(v-1):
u1,u2=In()
adj[u1].add(u2)
adj[u2].add(u1)
return adj,v
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*1000001
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range(1000001):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
mx=10**7
spf=[mx]*(mx+1)
def SPF():
spf[1]=1
for i in range(2,mx+1):
if spf[i]==mx:
spf[i]=i
for j in range(i*i,mx+1,i):
if i<spf[j]:
spf[j]=i
return
#####################################################################################
mod=10**9+7
def solve():
n=int(input())
li = []
for i in range(n):
x=[]
for j in range(n):
if i%2:
val=pow(2,i+j)
else:
val=0
x.append(val)
li.append(x)
for ele in li:
print(*ele,flush=True)
stout.flush()
q=II()
for _ in range(q):
v=II()
v=v>>1
print(1,1,flush=True)
x,y=1,1
h=0
while v:
if h%2==0:
if v&1:
x+=1
h+=1
else:
y+=1
else:
if v&1:
y+=1
else:
x+=1
h+=1
v=v>>1
print(x,y,flush=True)
stdout.flush()
return
t=1
for i in range(t):
solve()
#######
#
#
####### # # # #### # # #
# # # # # # # # # # #
# #### # # #### #### # #
###### # # #### # # # # #
```
No
| 103,068 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
r=0
for i in range(1,n):
r=r+int((k-l[i])/l[0])
print(r)
```
| 103,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
'''Author- Akshit Monga'''
t=int(input())
for _ in range(t):
n,k=map(int,input().split())
arr=[int(x) for x in input().split()]
m=min(arr)
ans=0
for i in arr:
ans+=(k-i)//m
print(ans-(k-m)//m)
```
| 103,070 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Sep 19 16:36:57 2020
@author: lakne
"""
t = int(input())
answers = []
for _ in range(t):
x = 0
nk = input().split()
a = input().split()
a_sorted = []
for i in range(int(nk[0])):
a_sorted.append(int(a[i]))
a_sorted = sorted(a_sorted)
for j in range(int(nk[0])-1):
x += int((int(nk[1])-a_sorted[-1-j])/a_sorted[0])
answers.append(int(x))
for k in range(t):
print(answers[k])
```
| 103,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
a=sorted(list(map(int,input().split())))
x=a[0]
a.pop(0)
ans=0
for i in range(n-1):
x1=k-a[i]
ans+=(x1//x)
print(ans)
```
| 103,072 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
# 1417A.py
for _ in range(int(input())):
n,k = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
spell = 0
for i in range(1,n):
a[i] = a[i]-k
if a[i]<0:
a[i]=-1*a[i]
spell += (a[i]//a[0])
else:
break
print(spell)
```
| 103,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
ans = 0
ref = l[0]
for i in range(1,n):
if l[i]>=k:
continue
dif = k - l[i]
ans += dif//ref
print(ans)
```
| 103,074 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
a = list(map(int, input().split()))
ma = min(a)
ind = a.index(ma)
ans = 0
for i in range(n):
if i == ind:
continue
m = (k - a[i]) // ma
a[i] += m * ma
ans += m
print(ans)
```
| 103,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n,k=[int(i) for i in input().split()]
l=[int(i) for i in input().split()]
l.sort()
mn=l[0]
ans=0
for i in range(n-1, 0,-1):
if(l[i]<k):
d=k-l[i]
ans+=int(d/mn)
print(ans)
```
| 103,076 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
for ad in range(int(input())):
n,k=list(map(int,input().split()))
l=list(map(int,input().split()))
l.sort()
x=l[0]
ans=0
for i in range(n-1,0,-1):
y=(k-l[i])/x
ans+=int(y)
print(ans)
```
Yes
| 103,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
import itertools as it
for _ in range(int(input())):
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
counter = 0
for i in range(1, len(a)):
counter += (k - a[i])//a[0];
print(counter)
```
Yes
| 103,078 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
from sys import stdin,stdout
#from collections import defaultdict
st=lambda:list(stdin.readline().strip())
li=lambda:list(map(int,stdin.readline().split()))
mp=lambda:map(int,stdin.readline().split())
inp=lambda:int(stdin.readline())
pr=lambda n: stdout.write(str(n)+"\n")
#from bisect import bisect_left,bisect_right
mod=1000000007
INF=float('inf')
def solve():
n,k=mp()
l=li()
l.sort()
ans=0
a=l[0]
for i in range(1,n):
if l[i]<=k:
ans+=(k-l[i])//a
pr(ans)
for _ in range(inp()):
solve()
```
Yes
| 103,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
import sys
def solution():
get_input = sys.stdin.readline
t = int(get_input().strip())
for i in range(t):
n, k = list(map(int, get_input().strip().split()))
board = list(map(int, get_input().strip().split()))
board.sort()
mb = board[0]
total = 0
for j in range(1, n):
total += ((k - board[j]) // mb)
print(total)
solution()
```
Yes
| 103,080 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
print((k - a[1]) // a[0])
```
No
| 103,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
for i in range(int(input())):
n,k=map(int,input().split())
l=sorted(list(map(int,input().split())))
maxopr=0
m=min(l)
for i in l:
if i!=m:
if (k-i)%m==0:
maxopr+=(k-i)//m-1
else:
maxopr+=(k-i)//m
else:
for i in l:
if i!=m:
m=i
break
if (k-i)%m==0:
maxopr+=(k-i)//m-1
else:
maxopr+=(k-i)//m
m=min(l)
c=0
for i in l:
if (k-i)%m==0:
c=1
break
if c==0:
print(maxopr)
else:
print(maxopr+1)
```
No
| 103,082 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
for _ in range(int(input())):
n,t=map(int,input().split())
arr=list(map(int,input().split()))
ans=0
arr.sort()
for j in range(1,n):
ans+=((t-arr[j])//arr[0])
print(ans//arr[0])
```
No
| 103,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
— Hey folks, how do you like this problem?
— That'll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows:
1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j.
2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed.
BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input
The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.
Each test case consists of two lines:
* the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4);
* the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k).
It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.
Output
For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example
Input
3
2 2
1 1
3 5
1 2 3
3 7
3 2 2
Output
1
5
4
Note
In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.
Submitted Solution:
```
from sys import stdin
input=lambda:stdin.readline().strip()
for _ in range(int(input())):
n,k=[int(i) for i in input().split()]
lst=[int(i) for i in input().split()]
min1=min(lst)
lst.remove(min1)
nd=0
lst2=[]
for i in lst:
nk=(k-i)//min1
lst2.append(i+(nk)*((k-i)//min1))
nd+=nk
min2=min(lst2)
nd+=(k-min1)//min2
print(nd)
```
No
| 103,084 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
n, m, k = map(int, input().split())
# a = []
# for _ in range(n):
# a.append(list(map(int, input().split())))
dp = [float("-inf") for i in range(k)]
dp[0] = 0
sz = m // 2
for _ in range(n):
a = list(map(int, input().split()))
pre = dp
dp = [[float("-inf") for _ in range(k)] for _ in range(sz + 1)]
dp[0] = pre
for x in a:
for i in range(sz, 0, -1):
for j in range(k):
dp[i][j] = max(dp[i][j], dp[i - 1][((j - x) % k + k) % k] + x)
for i in range(1, sz + 1):
for j in range(k):
dp[0][j]=max(dp[0][j], dp[i][j])
dp = dp[0]
print(dp[0])
```
| 103,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def IND(i,j):
return i*K + j
def ind(i,j,k):
return i*(m//2+1)*K + j*K + k
n,m,K = inpl()
DP = [0] * (K*(n+1))
for i in range(n):
dp = [0] * (K*(m//2+1)*(m+1))
a = inpl()
for j,x in enumerate(a):
for cnt in range(m//2):
for mo in range(K):
dp[ind(j+1,cnt,mo)] = dp[ind(j,cnt,mo)]
for cnt in range(m//2):
for mo in range(K):
now = dp[ind(j,cnt,mo)]
dp[ind(j+1,cnt+1,(now+x)%K)] = max(dp[ind(j,cnt+1,(now+x)%K)], dp[ind(j+1,cnt+1,(now+x)%K)], now+x)
for mo in range(K):
mx = 0
for j in range(m+1):
for cnt in range(m//2+1):
mx = max(dp[ind(j,cnt,mo)], mx)
for mm in range(K):
if DP[IND(i,mm)]%K != mm: continue
now = DP[IND(i,mm)]
DP[IND(i+1,(now+mx)%K)] = max(DP[IND(i,(now+mx)%K)],DP[IND(i+1,(now+mx)%K)], now+mx)
print(DP[IND(n,0)])
```
| 103,086 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
from math import inf as _inf
import sys as _sys
_NEGATIVE_INF = -_inf
def main():
n, m, k = _read_ints()
a = tuple(tuple(_read_ints()) for i_row in range(n))
result = find_max_sum(a, k)
print(result)
def _read_line():
result = _sys.stdin.readline()
assert result[-1] == "\n"
return result[:-1]
def _read_ints():
return map(int, _read_line().split())
def find_max_sum(matrix, k):
selections_n = len(matrix[0]) // 2
rows_maximums = tuple(map(lambda row: _find_seq_maximums(row, selections_n, k), matrix))
cache = [[None] * k for i_row in range(len(matrix) + 1)]
cache[len(matrix)][0] = 0
for remainder_need in range(1, k):
cache[len(matrix)][remainder_need] = _NEGATIVE_INF
def _f_matrix(i_row, remainder_need):
cached_value = cache[i_row][remainder_need]
if cached_value is not None:
return cached_value
result = _NEGATIVE_INF
for remainder, maximum in rows_maximums[i_row].items():
potential_result = maximum + _f_matrix(i_row + 1, (remainder_need - remainder) % k)
if potential_result > result:
result = potential_result
cache[i_row][remainder_need] = result
return result
return _f_matrix(0, 0)
def _find_seq_maximums(seq, selections_n, k):
cache = [None] * (len(seq) + 1)
for i in range(len(cache)):
cache_for_i = [None] * k
for remainder_need in range(k):
cache_for_i[remainder_need] = [None] * (selections_n + 1)
cache[i] = cache_for_i
for i in range(len(seq) + 1):
cache[i][0][0] = 0
for remainder_need in range(1, k):
cache[i][remainder_need][0] = _NEGATIVE_INF
for can_select in range(selections_n + 1):
cache[len(seq)][0][can_select] = 0
for remainder_need in range(1, k):
cache[len(seq)][remainder_need][can_select] = _NEGATIVE_INF
for i in reversed(range(len(seq))):
for remainder_need in range(k):
for can_select in range(1, selections_n + 1):
skip_result = cache[i + 1][remainder_need][can_select]
take_result = seq[i] + cache[i + 1][(remainder_need - seq[i]) % k][can_select - 1]
if take_result > skip_result:
result = take_result
else:
result = skip_result
cache[i][remainder_need][can_select] = result
result = dict()
for remainder in range(k):
maximum = cache[0][remainder][selections_n]
if maximum > _NEGATIVE_INF:
result[remainder] = maximum
return result
if __name__ == '__main__':
main()
```
| 103,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
n, m, k = map(int, input().split())
l = m // 2
p = [-1] * k
p[0] = 0
for _ in range(n):
t = [[-1] * (l + 1) for i in range(k)]
t[0][0] = 0
for q in map(int, input().split()):
for j in range(l - 1, -1, -1):
for i in range(k):
if t[i][j] != -1:
d = (i + q) % k
if t[d][j + 1] < t[i][j] + q:
t[d][j + 1] = t[i][j] + q
r = [-1] * k
for i in range(k):
q = max(t[i])
if q == -1:
continue
for j in range(k):
if p[j] == -1:
continue
d = (i + j) % k
if r[d] < p[j] + q:
r[d] = p[j] + q
p = r
print(p[0])
```
| 103,088 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
n, m, k = getints()
m //= 2
dp = [int(-1e9)] * k
dp[0] = 0
for _ in range(n):
tmp = [[int(-1e9)] * (m + 1) for _ in range(k)]
tmp[0][0] = 0
for a in getints():
for x in range(m - 1, -1, -1):
for y in range(k):
col = (y + a) % k
tmp[col][x + 1] = max(tmp[col][x + 1], tmp[y][x] + a)
ndp = [int(-1e9)] * k
for x in range(k):
tmp[x][0] = max(tmp[x])
for y in range(k):
nxt = (x + y) % k
ndp[nxt] = max(ndp[nxt], dp[y] + tmp[x][0])
dp = ndp
print(dp[0])
# 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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def getint(): return int(input())
def getints(): return list(map(int, input().split()))
def getint1(): return list(map(lambda x: int(x) - 1, input().split()))
# endregion
if __name__ == "__main__":
main()
```
| 103,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for k in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for k in range(c)] for k in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10
N, M, K = MAP()
MX = 70
dp = list4d(N+1, M+1, M//2+1, K, -1)
dp[0][0][0][0] = 0
for i in range(N):
row = LIST()
for j in range(M):
for k in range(M//2+1):
for l in range(K):
if dp[i][j][k][l] != -1:
dp[i][j+1][k][l] = max(dp[i][j+1][k][l], dp[i][j][k][l])
if k+1 <= M//2:
dp[i][j+1][k+1][(l+row[j])%K] = max(dp[i][j+1][k+1][(l+row[j])%K], dp[i][j][k][l] + row[j])
for k in range(M//2+1):
for l in range(K):
dp[i+1][0][0][l] = max(dp[i+1][0][0][l], dp[i][M][k][l])
ans = dp[N][0][0][0]
print(ans)
```
| 103,090 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
n, m, k = map(int, input().split())
dp = [-10000000000 for i in range(k)]
dp[0] = 0
sz = m//2
mat = []
for i in range(n):
mat.append(list(map(int,input().split())))
for a in mat:
pre = dp
dp = [[-10000000000 for i in range(k)] for j in range(sz+1)]
dp[0] = pre
for x in a:
for i in range(sz,0,-1):
for j in range(k):
dp[i][j] = max(dp[i][j], dp[i-1][((j-x)%k+k)%k] + x)
for i in range(1,sz+1):
for j in range(k):
dp[0][j]=max(dp[0][j], dp[i][j])
dp = dp[0]
print(dp[0])
```
| 103,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Tags: dp
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,m,k = map(int,input().split())
arr = [list(map(int,input().split())) for _ in range(n)]
dp = [[[0]*k for _ in range(m//2+1)]for _ in range(n)]
for i in range(n):
for j in range(m):
for r in range(min(m//2+1,j+2)-1,0,-1):
for x in range(k):
y = arr[i][j]+dp[i][r-1][x]
dp[i][r][y%k] = max(dp[i][r][y%k],y)
if i != n-1:
for xx in range(k):
for yy in range(m//2+1):
dp[i+1][0][xx] = max(dp[i+1][0][xx],dp[i][yy][xx])
a = 0
for i in range(m//2+1):
a = max(a,dp[-1][i][0])
print(a)
#Fast IO Region
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")
if __name__ == '__main__':
main()
```
| 103,092 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Submitted Solution:
```
n, m, k = map(int, input().split())
# a = []
# for _ in range(n):
# a.append(list(map(int, input().split())))
pre = [float("-inf") for i in range(k)]
pre[0] = 0
sz = m // 2
for _ in range(n):
a = list(map(int, input().split()))
dp = [[float("-inf") for _ in range(k)] for _ in range(sz + 1)]
dp[0] = pre
for x in a:
for i in range(sz, 0, -1):
for j in range(k):
dp[i][j] = max(dp[i][j], dp[i - 1][(j - x) % k] + x)
for i in range(1, sz + 1):
for j in range(k):
dp[0][j] = max(dp[0][j], dp[i][j])
pre = dp[0]
print(pre[0])
```
Yes
| 103,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Submitted Solution:
```
import copy
def get_h(data, n, k):
h = []
for _ in range(k):
h.append([0] * (n + 1))
hh = copy.deepcopy(h)
for v in data:
for j in range(1, n + 1):
for i in range(k):
t = (h[i][j - 1] + v) % k
tt = (i + v) % k
if t == tt:
hh[t][j] = max(hh[t][j], h[i][j - 1] + v)
hh, h = h, hh
for j in range(0, n + 1):
for i in range(k):
hh[i][j] = h[i][j]
#print(hh, h)
return [max(t) for t in h]
n, m, k = list(map(int, input().split()))
bh = [0] * k
tbh = [0] * k
for _ in range(n):
data = list(map(int, input().split()))
h = get_h(data, m // 2, k)
for t in range(k):
for i in range(k):
j = (t - i + k) % k
if (bh[i] + h[j]) % k == t:
tbh[t] = max(tbh[t], bh[i] + h[j])
bh, tbh = tbh, bh
for i in range(k):
tbh[i] = bh[i]
print(bh[0])
```
Yes
| 103,094 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
n, m, k = getints()
m //= 2
dp = [int(-1e9)] * k
dp[0] = 0
for _ in range(n):
tmp = [[int(-1e9)] * k for _ in range(m + 1)]
tmp[0][0] = 0
for a in getints():
for x in range(m - 1, -1, -1):
for y in range(k):
col = (y + a) % k
tmp[x + 1][col] = max(tmp[x + 1][col], tmp[x][y] + a)
ndp = [int(-1e9)] * k
for i in range(m + 1):
for x in range(k):
for y in range(k):
nxt = (x + y) % k
ndp[nxt] = max(ndp[nxt], dp[y] + tmp[i][x])
dp = ndp
print(dp[0])
# 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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def getint(): return int(input())
def getints(): return list(map(int, input().split()))
def getint1(): return list(map(lambda x: int(x) - 1, input().split()))
# endregion
if __name__ == "__main__":
main()
```
Yes
| 103,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Submitted Solution:
```
INF = int(1e9)
n, m, k = map(int, input().split())
a = [[] for i in range(n)]
for i in range(n):
for j in map(int, input().split()):
a[i].append(j)
dp = [[[-INF for q in range(k)] for p in range(m+1)] for j in range(m)]
best = [-INF for q in range(k)]
best[0] = 0
for i in range(n):
for j in range(m):
for p in range(m+1):
for q in range(k):
if p > j+1 or p > m//2:
dp[j][p][q] = -INF
elif j == 0:
dp[j][p][q] = max(a[i][j] + best[(q-a[i][j]+k)%k] if p==1 else -INF, best[q])
else:
dp[j][p][q] = max(a[i][j] + dp[j-1][p-1][(q-a[i][j]+k)%k], dp[j-1][p][q])
for q in range(k):
for p in range(m+1):
best[q] = max(best[q], dp[m-1][p][q])
print(best[0])
```
Yes
| 103,096 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Submitted Solution:
```
from itertools import permutations
#from fractions import Fraction
from collections import defaultdict
from math import*
import os
import sys
from io import BytesIO, IOBase
from heapq import nlargest
from bisect import*
import copy
import itertools
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
#-------------above part copied-----------------------
n,m,test=map(int,input().split())
arr=[]
ans=0
for i in range(n):
brr=list(map(int,input().split()))
brr.sort(reverse=True)
arr.append(brr)
for i in range(m//2):
ans+=brr[i]
if ans%test==0:
print(ans)
else:
temp=0
for i in range(n):
for j in range(m//2,m):
for k in range((m//2)-1,-1,-1):
if (ans+arr[i][j]-arr[i][k])%test==0:
val = ans+arr[i][j]-arr[i][k]
if val>temp:
temp=val
break
break
print(temp)
```
No
| 103,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Submitted Solution:
```
n,m,k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(n)]
dp1 = []
for i in range(n):
B = A[i]
temp = [[-1]*(k) for _ in range(m//2+1)]
temp[0][0] = 0
for b in B:
for i in reversed(range(m//2+1)):
if i == m//2:
continue
for j in range(k):
if temp[i][j] != -1:
temp[i+1][(b+j)%k] = max(temp[i+1][(b+j)%k], temp[i][j]+b)
temp2 = [-1]*k
for i in range(m//2+1):
for j in range(k):
temp2[j] = max(temp2[j], temp[i][j])
dp1.append(temp2)
#print(dp1)
dp2 = [[-1]*k for i in range(n+1)]
dp2[0][0] = 0
for i in range(n):
for j in range(k):
for l in range(k):
dp2[i+1][(j+l)%k] = max(dp2[i+1][(j+l)%k], dp2[i][l]+dp1[i][j])
ans = dp2[n][0]
print(ans)
```
No
| 103,098 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a matrix a of size n × m consisting of integers.
You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.
In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.
Note that you can choose zero elements (and the sum of such set is 0).
Input
The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).
Output
Print one integer — the maximum sum divisible by k you can obtain.
Examples
Input
3 4 3
1 2 3 4
5 2 2 2
7 1 1 4
Output
24
Input
5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6
Output
56
Note
In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Oct 20 21:47:44 2020
@author: Dark Soul
"""
[n,m,k]=list(map(int,input().split()))
mn=[]
ans=0
for i in range(n):
arr=list(map(int,input().split()))
arr.sort(reverse=True)
mn.append(arr)
for j in range(m//2):
ans+=arr[j]
if ans%k==0:
print(ans)
else:
sol=-1
for i in range(n):
dic={}
y=mn[i]
now=[]
nans=0
for j in range(m//2,m):
now.append(y[j])
if len(now)==0:
continue
for j in now:
try:
dic[i]=1
except:
dic[i]=1
for j in now:
if (ans-j)%k==0:
nans=max(nans,ans-j)
try:
dic[k]
nans=max(nans,ans+k)
except:
continue
else:
mynum=ans-j
dorkar=k-(ans%k)
try:
dic[dorkar]
nans=max(nans,ans+dorkar)
except:
continue
sol=max(sol,nans)
print(sol)
```
No
| 103,099 |
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