message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
from collections import Counter
c = Counter(input())
print("Yes" if c['o'] == 0 or not c['-']%c['o'] else "No")
``` | instruction | 0 | 99,231 | 7 | 198,462 |
Yes | output | 1 | 99,231 | 7 | 198,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
s=input()
n=len(s)
count=0
for i in range(n):
if s[i]=="o":
count+=1
yo=n-count
if count==0:
print("YES")
elif (yo%count==0):
print("YES")
else:
print("NO")
``` | instruction | 0 | 99,232 | 7 | 198,464 |
Yes | output | 1 | 99,232 | 7 | 198,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
def solve(linha, perolas):
if perolas == 0:
return True
else:
return linha % perolas == 0
def main():
s = input()
x, y = 0, 0
for c in s:
if c == '-':
x += 1
else:
y += 1
if solve(x, y):
print('YES')
else:
print('NO')
if __name__ == "__main__":
main()
``` | instruction | 0 | 99,233 | 7 | 198,466 |
Yes | output | 1 | 99,233 | 7 | 198,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
def main():
necklace = input()
n_pearls = necklace.count('o')
n_links = len(necklace) - n_pearls
print('YES' if n_pearls == 0 or n_links % n_pearls == 0 else 'NO')
if __name__ == '__main__':
main()
``` | instruction | 0 | 99,234 | 7 | 198,468 |
Yes | output | 1 | 99,234 | 7 | 198,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
n=input()
l=len(n)
x=n.count('o')
if l%(x+1) and x!=l:
print("NO")
else:
print("YES")
``` | instruction | 0 | 99,235 | 7 | 198,470 |
No | output | 1 | 99,235 | 7 | 198,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
l = list(input())
c = len(l)
if l.count('-')%2==0:
print("YES")
elif l.count('o')==0:
print("YES")
elif l.count('o')==c:
print("YES")
elif l.count('-')%2==1 and l.count('o')%2==1:
print("YES")
elif l.count('o')%2==1 and l.count('-')%2==0:
print("NO")
else:
print("NO")
``` | instruction | 0 | 99,236 | 7 | 198,472 |
No | output | 1 | 99,236 | 7 | 198,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
string = input()
link = string.count('-')
pearl = len(string) - link
if pearl > 0 and link % pearl == 0:
# if link == 0 or pearl == 1, link % pearl == 0
print("YES")
else:
print("NO")
``` | instruction | 0 | 99,237 | 7 | 198,474 |
No | output | 1 | 99,237 | 7 | 198,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
<image>
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input
The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Examples
Input
<span class="tex-font-style-tt">-o-o--</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---</span>
Output
YES
Input
<span class="tex-font-style-tt">-o---o-</span>
Output
NO
Input
ooo
Output
YES
Submitted Solution:
```
print("YES")
``` | instruction | 0 | 99,238 | 7 | 198,476 |
No | output | 1 | 99,238 | 7 | 198,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,594 | 7 | 199,188 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import math
def f(n):
c=[]
d=[]
i = 1
while i <= math.sqrt(n):
if (n % i == 0):
if (n // i == i):
c.append(i)
else:
c.append(i)
d.append(n//i)
i+=1
c+=d[::-1]
return c
t=int(input())
for i in range(t):
n,k=map(int,input().split())
s=input()
b=f(k)
ans=0
c=[0]*(26)
for j in range(n):
c[ord(s[j])-97]+=1
k=[]
for j in range(26):
p=c[j]
i=1
while((p//i)!=0):
k.append(p//i)
i+=1
k.sort(reverse=True)
j=0
r=len(k)
while(j<len(b)):
p=b[j]
if (p-1)<r:
ans=max(ans,p*k[p-1])
j+=1
print(ans)
``` | output | 1 | 99,594 | 7 | 199,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,595 | 7 | 199,190 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
def binary_search(item,start,end,l):
if (start>end):
return max(end,0)
if (start>len(l)-1):
return (len(l)-1)
mid=(start+end)//2
if l[mid]>item:
return binary_search(item,start,mid-1,l)
elif(l[mid]<item):
return binary_search(item,mid+1,end,l)
else:
return mid
def f():
n,k=map(int,input().split(" "))
multiplesofk=[]
for i in range(1,k+1):
if (k%i==0):
multiplesofk.append(i)
string=input()
counter=[0]*200
for i in range(0,len(string)):
counter[ord(string[i])]+=1
listofcount=[]
for i in range(ord('a'),ord('z')+1):
if counter[i]!=0:
listofcount.append(counter[i])
maximum=0
for i in range(1,max(listofcount)+1):
s=0
for j in range(0,len(listofcount)):
s+=listofcount[j]//i
element=binary_search(s,0,len(multiplesofk)-1,multiplesofk)
maximum=max(maximum,multiplesofk[element]*i)
print (maximum)
for i in range(int(input())):
f()
'''200 559
luzrgywfemprizaxktszmugkfdakyjfdwxrfcjouklnxjzypgdlxkyplaornwpktkedqimlvzpcaivmtavwskvelctnvyziytxudzxudfueyukgagjpwkltpfmowhcwjdfxzzjziuaegmwzsisthasicflftxfkfukywxndsdfdbjwhncmjfmbevlkybpxesyrgnssow
'''
``` | output | 1 | 99,595 | 7 | 199,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,596 | 7 | 199,192 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
letters = 26
for t in range(int(input())):
n, k = map(int, input().split(' '))
s = input()
counts = [0] * (letters + 1)
for c in s:
counts[ord(c) - ord('a')] += 1
counts.sort(reverse=True)
zeroi = counts.index(0)
if zeroi >= 0:
counts = counts[:zeroi]
#print('s:', s)
#print('counts:', counts)
ans = 1
for d in range(1, min(k,n)+1):
if k % d != 0:
continue
l = 1
r = n // d
while l < r:
m = (l + r + 1) // 2
inGroup = 0
for count in counts:
inGroup += count // m
if inGroup >= d:
l = m
else:
r = m - 1
#print(f"d={d} groups={l} len={l*d}")
ans = max(ans, l*d)
print(ans)
``` | output | 1 | 99,596 | 7 | 199,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,597 | 7 | 199,194 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
import bisect
import collections
from collections import deque
import heapq
from collections import defaultdict
t = int(input())
for f in range(t):
n,k = map(int,input().split())
div = []
for i in range(1,2001):
if k%i == 0:
div.append(i)
s = list(input().rstrip())
c = collections.Counter(s)
a = c.most_common()
freq = []
for i in a:
freq.append(i[1])
ans = 0
for i in div:
j = 1
tmp = 0
while j*i <=n:
memo = 0
for l in range(len(freq)):
memo += freq[l]//j
if memo >= i:
tmp = j*i
j += 1
else:
break
ans = max(ans,tmp)
print(ans)
``` | output | 1 | 99,597 | 7 | 199,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,598 | 7 | 199,196 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import os
### START FAST IO ###
inp = os.read(0, int(1e7)).split()
inp_pos = -1
def cin():
global inp_pos
inp_pos += 1
return inp[inp_pos]
# cout: os.write(1, "\n".encode().join(res))
#### END FAST IO ####
from itertools import groupby
from math import gcd
T = int(cin())
res = []
while T:
T -= 1
n = int(cin())
k = int(cin())
a = [len(list(g)) for k, g in groupby(sorted(cin()))]
ans = 0
for v in range(1, n+1):
d = gcd(v, k)
f = v//d
if sum([x//f for x in a]) >= d:
ans = v
res.append(str(ans).encode())
os.write(1, "\n".encode().join(res))
``` | output | 1 | 99,598 | 7 | 199,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,599 | 7 | 199,198 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import atexit
import io
import sys
import math
from collections import defaultdict,Counter
# _INPUT_LINES = sys.stdin.read().splitlines()
# input = iter(_INPUT_LINES).__next__
# _OUTPUT_BUFFER = io.StringIO()
# sys.stdout = _OUTPUT_BUFFER
# @atexit.register
# def write():
# sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
# sys.stdout=open("CP2/output.txt",'w')
# sys.stdin=open("CP2/input.txt",'r')
# m=pow(10,9)+7
t=int(input())
for i in range(t):
n,k=map(int,input().split())
s=input()
if k%n==0:
print(n)
continue
c2=list(Counter(s).values())
# print(c)
for j in range(n,0,-1):
c=c2[:]
d=defaultdict(list)
for kk in range(j):
d[kk]=(kk+k)%j
visit=[0]*j
l=[]
for kk in range(j):
if visit[kk]==0:
c1=1
visit[kk]=1
p=d[kk]
while visit[p]==0:
c1+=1
visit[p]=1
p=d[p]
l.append(c1)
l.sort(reverse=True)
# print(l)
for kk in l:
m=max(c)
if m>=kk:
ind=c.index(m)
c[ind]-=kk
else:
break
else:
print(j)
break
``` | output | 1 | 99,599 | 7 | 199,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,600 | 7 | 199,200 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def li():
return [int(x) for x in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def bo(i):
return ord(i)-ord('a')
from copy import *
from bisect import *
from itertools import permutations
def can(mid,k):
if k==0:
return True
#print(d,mid,k)
if mid %k!=0 and max(d)<mid:
return False
elif mid%k!=0:
return True
c=i=0
while c<k:
#print(c,b)
if b[i]<mid//k:
return False
b[i]-=mid//k
c+=1
i=(i+1)%p
if b[i]<mid//k:
i=0
#print(b,mid)
return True
tt=fi()
while tt>0:
tt-=1
n,k=mi()
#k%=n
s=si()
d=[0 for i in range(26)]
for i in s:
d[bo(i)]+=1
p=k
#print("K",k)
if k==0:
print(n)
continue
div=[]
while p<=n:
div.append(p)
p+=k
#print(div)
div=sorted(div)
l=0
r=len(div)-1
d.sort(reverse=True)
if d.count(0)>0:
d=d[:d.index(0)]
p=len(d)
#print(d)
ans=max(d)
for mid in range(1,n+1):
b=deepcopy(d)
if can(mid,gcd(k,mid)):
ans=mid
print(ans)
``` | output | 1 | 99,600 | 7 | 199,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo". | instruction | 0 | 99,601 | 7 | 199,202 |
Tags: brute force, dfs and similar, dp, graphs, greedy, number theory
Correct Solution:
```
import sys
input = sys.stdin.readline
from math import gcd
t=int(input())
for tests in range(t):
n,k=map(int,input().split())
S=input().strip()
LIST=[0]*26
for s in S:
LIST[ord(s)-97]+=1
for x in range(n,0,-1):
GCD=gcd(x,k)
p=x//GCD
SUM=0
for l in LIST:
if l>=p:
SUM+=l//p*p
if SUM>=x:
print(x)
break
#print(S)
``` | output | 1 | 99,601 | 7 | 199,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
import math
num_tests = int(input())
def calc_num_least(letters, x):
count = 0
for letter in letters:
if letters[letter] >= x:
count += letters[letter] // x
return count
def solve(n, k, letters):
best = 0
for i in range(1, n+1):
j = calc_num_least(letters, i)
for x in reversed(range(1, j+1)):
if k % x == 0:
best = max(best, x * i)
break
return best
for test in range(num_tests):
n, k = [int(x) for x in input().split(" ")]
s = input()
letters = {}
for i in s:
letters[i] = letters.get(i, 0) + 1
print(solve(n, k, letters))
``` | instruction | 0 | 99,602 | 7 | 199,204 |
Yes | output | 1 | 99,602 | 7 | 199,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
import sys
from math import gcd
from bisect import bisect_left as br
from collections import Counter as CO
from collections import defaultdict as dd
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n,kk=map(int,input().split())
word=list(map(str,input().strip()))
df=dd(int)
fact=[]
for k in word:
df[k]+=1
#print(df)
ans=0
for i in range(1,n+1):
maxi=0
x=0
co=0
for j in df:
if(df[j]//i!=0):
maxi+=df[j]//i
#rint(maxi)
if(maxi==0):
break
for j in range(maxi,0,-1):
if(kk%j==0 ):
maxi=j
break
ans=max(ans,maxi*i)
#print(i,maxi)
print(ans)
``` | instruction | 0 | 99,603 | 7 | 199,206 |
Yes | output | 1 | 99,603 | 7 | 199,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
"""n=int(input())
a=list(map(int,input().split()))
ct=[0]*20
for x in a:
for sf in range(20):
ct[sf]+=(x>>sf)&1
ans=[0]*n
for i in range(20):
for j in range(ct[i]):
print(ans[j],1<<i,ans[j] | 1<<i)
ans[j]|=1<<i
print(sum([x**2 for x in ans]))
"""
"""k=int(input())
s='codeforces'
i,sm=0,1
l=[1]*10
while sm<k:
sm//=l[i]
print(sm,l[i])
l[i]+=1
print(l[i])
sm*=l[i]
print(sm)
i=(i+1)%10
print(l)
ans=[s[i]*l[i] for i in range(10)]
print(''.join(ans))"""
"""T =int(input())
for _ in range(T):
a=input()
if len(a)==2:
print(a)
else:
for i in range(0,len(a),2):
print(a[i],end="")
print(a[-1])"""
"""T = int(input())
for _ in range(T):
n = int(input())
arr = list(map(int,input().split()))
a=b=0
for i in range(n):
if i%2 != arr[i]%2:
if i%2==0:
a+=1
else:
b+=1
if a==b:
print(a)
else:
print(-1)"""
"""T= int(input())
for _ in range(T):
n,k = list(map(int,input().split()))
a = input()
e='0'*k+a+'0'*k
f = k*2+1
z=ct=0
for i in e.split('1'):
ct+=max((len(i)-k)//(k+1),0)
print(ct)"""
"""for _ in range(int(input())):
arr = ['#']+sorted(list(input()))
n = int(input())
num = [-1]+[int(i) for i in input().split()]
ans = ['#']*(n+1)
while 0 in num:
zero = []
for i in range(1,n+1):
if num[i]==0:
zero.append(i)
num[i]=-1
#print("zero",zero)
while arr.count(arr[-1])<len(zero):
p = arr[-1]
while p==arr[-1]:
arr.pop()
#print("arr",arr)
p=0
for i in zero:
p=ans[i]=arr[-1]
#print("ans",ans)
while arr[-1]==p:
arr.pop()
for i in range(1,n+1):
if ans[i]=='#':
for j in zero:
num[i]-=abs(j-i)
#print("num",num)
print(''.join(ans[1:]))"""
import math
import collections
for _ in range(int(input())):
n,k = map(int, input().split())
s = input()
ctr = collections.Counter(s)
ans = 0
for i in range(1,n+1):
g = math.gcd(i, k)
l = i // g
#print(i,k,g,l)
cnt = sum([(v//l)*l for v in ctr.values()])
#print(cnt)
if cnt >= i:
ans = i
print(ans)
``` | instruction | 0 | 99,604 | 7 | 199,208 |
Yes | output | 1 | 99,604 | 7 | 199,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
def gcd(a, b):
if a < b:
return gcd(b, a)
if b == 0:
return a
return gcd(b, a%b)
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
if s[0] * n == s:
print(n)
continue
cnt = {}
for c in s:
if c not in cnt:
cnt[c] = 0
cnt[c] += 1
cnt = [cnt[c] for c in cnt]
ans = 1
for x in range(2,n+1):
tempK = k % x
if tempK == 0:
ans = max(ans, x)
continue
mod = x % tempK
if mod != 0:
mod = -(mod - tempK)
times = tempK // gcd(tempK, mod)
needed = x*times // tempK
usedBeads = 0
for num in cnt:
usedBeads += num - num % needed
if usedBeads >= x:
ans = max(ans, x)
print(ans)
``` | instruction | 0 | 99,605 | 7 | 199,210 |
Yes | output | 1 | 99,605 | 7 | 199,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
def ints():
return list(map(int,input().split()))
from collections import Counter
import math
def pf(n):
l = []
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return list(set(l))
for _ in range(int(input())):
n,k = ints()
s = input()
p = pf(k)
d = Counter(s)
d = sorted(d.values(),reverse = True)
if k==1:
print (n)
continue
m = 0
#print (d)
#print(p)
for x in p:
c = 0
for i in range(len(d)):
if d[i]<x:
break
q = d[i]//x
c = max(c,x*q*(i+1))
#print(c)
if c==0 and len(d)>=x:
c = x
m = max(m,c)
c = 0
if len(d)>=x:
c = d[x-1]
m = max(m,c*x)
#print (m)
#print ()
print(m)
``` | instruction | 0 | 99,606 | 7 | 199,212 |
No | output | 1 | 99,606 | 7 | 199,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
def good(x,n,k):
m=0
l=[]
for i in range(n+1):
for j in range(i+1,n+1):
a=x[i:j]
c=a[k:]+a[:k]
if a==c:
l.append(a)
m=len(a)
return(l)
for _ in range(int(input())):
n,k=map(int,input().split())
x=str(input())
print(good(x,n,k))
``` | instruction | 0 | 99,607 | 7 | 199,214 |
No | output | 1 | 99,607 | 7 | 199,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
import sys
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def find(cnt, k, h):
l = 0
r = 2001
while r-l > 1:
now = 0
c = (l + r) // 2
for j in cnt:
now += j//c
if now >= h:
l = c
else:
r = c
return l*h
def solve():
n, k = mints()
cnt = [0] * 26
for i in minp():
cnt[ord(i) - ord('a')] += 1
cnt.sort(reverse = True)
ans = 0
for i in range(1,k):
if k % i == 0:
ans = max(ans, find(cnt, k, i))
if k // i != i:
ans = max(ans, find(cnt, k, k // i))
print(ans)
for i in range(mint()):
solve()
``` | instruction | 0 | 99,608 | 7 | 199,216 |
No | output | 1 | 99,608 | 7 | 199,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The store sells n beads. The color of each bead is described by a lowercase letter of the English alphabet ("a"–"z"). You want to buy some beads to assemble a necklace from them.
A necklace is a set of beads connected in a circle.
For example, if the store sells beads "a", "b", "c", "a", "c", "c", then you can assemble the following necklaces (these are not all possible options):
<image>
And the following necklaces cannot be assembled from beads sold in the store:
<image> The first necklace cannot be assembled because it has three beads "a" (of the two available). The second necklace cannot be assembled because it contains a bead "d", which is not sold in the store.
We call a necklace k-beautiful if, when it is turned clockwise by k beads, the necklace remains unchanged. For example, here is a sequence of three turns of a necklace.
<image> As you can see, this necklace is, for example, 3-beautiful, 6-beautiful, 9-beautiful, and so on, but it is not 1-beautiful or 2-beautiful.
In particular, a necklace of length 1 is k-beautiful for any integer k. A necklace that consists of beads of the same color is also beautiful for any k.
You are given the integers n and k, and also the string s containing n lowercase letters of the English alphabet — each letter defines a bead in the store. You can buy any subset of beads and connect them in any order. Find the maximum length of a k-beautiful necklace you can assemble.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the test. Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ n, k ≤ 2000).
The second line of each test case contains the string s containing n lowercase English letters — the beads in the store.
It is guaranteed that the sum of n for all test cases does not exceed 2000.
Output
Output t answers to the test cases. Each answer is a positive integer — the maximum length of the k-beautiful necklace you can assemble.
Example
Input
6
6 3
abcbac
3 6
aaa
7 1000
abczgyo
5 4
ababa
20 10
aaebdbabdbbddaadaadc
20 5
ecbedececacbcbccbdec
Output
6
3
5
4
15
10
Note
The first test case is explained in the statement.
In the second test case, a 6-beautiful necklace can be assembled from all the letters.
In the third test case, a 1000-beautiful necklace can be assembled, for example, from beads "abzyo".
Submitted Solution:
```
primes=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999]
from bisect import bisect_left as bl
from math import sqrt,ceil
t=int(input())
for q in range(t):
n,k=map(int,input().split())
s=input()
d={}
for i in s:
if i in d:
d[i]+=1
else:
d[i]=1
x=len(d.values())
while x>1 and k%x!=0:
x-=1
v=min(d.values())
m=max(d.values())
p=v*x
q=[]
if m>1:
prime_pt=bl(primes,ceil(sqrt(k)))
for i in range(prime_pt+1):
t=float("INF")
ct=0
if k%primes[i]==0:
for j in list(d.values()):
if j>=primes[i]:
t=min(t,(j//primes[i])*primes[i])
ct+=1
if t and (t*ct)%primes[i]==0:
q.append(t*ct)
t=float("INF")
ct=0
for j in list(d.values()):
if j>=k:
t=(min(t,j//k)*k)
ct+=1
if t and (t*ct)%k==0:
q.append(t*ct)
fq=0
if q:
fq=max(q)
print(max(p,fq))
``` | instruction | 0 | 99,609 | 7 | 199,218 |
No | output | 1 | 99,609 | 7 | 199,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a grid with n rows and m columns. Every cell of the grid should be colored either blue or yellow.
A coloring of the grid is called stupid if every row has exactly one segment of blue cells and every column has exactly one segment of yellow cells.
In other words, every row must have at least one blue cell, and all blue cells in a row must be consecutive. Similarly, every column must have at least one yellow cell, and all yellow cells in a column must be consecutive.
<image> An example of a stupid coloring. <image> Examples of clever colorings. The first coloring is missing a blue cell in the second row, and the second coloring has two yellow segments in the second column.
How many stupid colorings of the grid are there? Two colorings are considered different if there is some cell that is colored differently.
Input
The only line contains two integers n, m (1≤ n, m≤ 2021).
Output
Output a single integer — the number of stupid colorings modulo 998244353.
Examples
Input
2 2
Output
2
Input
4 3
Output
294
Input
2020 2021
Output
50657649
Note
In the first test case, these are the only two stupid 2× 2 colorings.
<image> | instruction | 0 | 99,678 | 7 | 199,356 |
Tags: combinatorics, dp, math
Correct Solution:
```
M=998244353;N=4042
try:
import __pypy__
int_add=__pypy__.intop.int_add
int_sub=__pypy__.intop.int_sub
int_mul=__pypy__.intop.int_mul
def make_mod_mul(mod=M):
fmod_inv=1.0/mod
def mod_mul(a,b,c=0):
res=int_sub(
int_add(int_mul(a,b),c),
int_mul(mod,int(fmod_inv*a*b+fmod_inv*c)),
)
if res>=mod:return res-mod
elif res<0:return res+mod
else:return res
return mod_mul
mod_mul=make_mod_mul()
except:
def mod_mul(a,b):return(a*b)%M
def mod_add(a,b):
v=a+b
if v>=M:v-=M
if v<0:v+=M
return v
def mod_sum(a):
v=0
for i in a:v=mod_add(v,i)
return v
f1=[1]
for i in range(N):f1.append(mod_mul(f1[-1],i+1))
f2=[pow(f1[-1],M-2,M)]
for i in range(N):f2.append(mod_mul(f2[-1],N-i))
f2=f2[::-1]
C=lambda a,b:mod_mul(mod_mul(f1[a],f2[b]),f2[a-b])
A=lambda a,b,w:mod_mul(C(a+b,a),C(w+b-a-2,b-1))
def V(h,W,H):
s=p=0
for i in range(W-1):
p=mod_add(p,A(i,H-h,W));s=mod_add(s,mod_mul(p,A(W-2-i,h,W)))
return s
H,W=map(int,input().split())
Y=mod_sum(mod_mul(A(s,h,W),A(W-2-s,H-h,W))for s in range(W-1)for h in range(1,H))
X=mod_add(mod_sum(V(h,W,H)for h in range(1,H)),mod_sum(V(w,H,W)for w in range(1,W)))
print((X+X-Y-Y)%M)
``` | output | 1 | 99,678 | 7 | 199,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once at New Year Dima had a dream in which he was presented a fairy garland. A garland is a set of lamps, some pairs of which are connected by wires. Dima remembered that each two lamps in the garland were connected directly or indirectly via some wires. Furthermore, the number of wires was exactly one less than the number of lamps.
There was something unusual about the garland. Each lamp had its own brightness which depended on the temperature of the lamp. Temperatures could be positive, negative or zero. Dima has two friends, so he decided to share the garland with them. He wants to cut two different wires so that the garland breaks up into three parts. Each part of the garland should shine equally, i. e. the sums of lamps' temperatures should be equal in each of the parts. Of course, each of the parts should be non-empty, i. e. each part should contain at least one lamp.
<image>
Help Dima to find a suitable way to cut the garland, or determine that this is impossible.
While examining the garland, Dima lifted it up holding by one of the lamps. Thus, each of the lamps, except the one he is holding by, is now hanging on some wire. So, you should print two lamp ids as the answer which denote that Dima should cut the wires these lamps are hanging on. Of course, the lamp Dima is holding the garland by can't be included in the answer.
Input
The first line contains single integer n (3 ≤ n ≤ 106) — the number of lamps in the garland.
Then n lines follow. The i-th of them contain the information about the i-th lamp: the number lamp ai, it is hanging on (and 0, if is there is no such lamp), and its temperature ti ( - 100 ≤ ti ≤ 100). The lamps are numbered from 1 to n.
Output
If there is no solution, print -1.
Otherwise print two integers — the indexes of the lamps which mean Dima should cut the wires they are hanging on. If there are multiple answers, print any of them.
Examples
Input
6
2 4
0 5
4 2
2 1
1 1
4 2
Output
1 4
Input
6
2 4
0 6
4 2
2 1
1 1
4 2
Output
-1
Note
The garland and cuts scheme for the first example:
<image>
Submitted Solution:
```
""" This is a solution to the problem Garland on codeforces.com
Given a rooted tree with node weights (possibly negative),
determine the two edges to be cut such that the sum of the weights in the remaining three subtrees is equal.
For details, see
http://codeforces.com/problemset/problem/767/C
"""
from sys import stdin
import sys
sys.setrecursionlimit(20000)
# compute subtree sums for all nodes, let S = total weight
# find two subtrees A, B with weight 1/3 S (check for divisibility first)
# distinguish two cases:
# - A is subtree of B: then A has total weight 2/3 S, B has weight 1/3 S
# - A and B are not descendants of one another
class RootedTree:
def __init__(self, n):
self._adj = {}
for i in range(n + 1):
self._adj[i] = []
def __repr__(self):
return str(self._adj)
@property
def root(self):
if len(self._adj[0]) == 0:
raise ValueError("Root not set.")
return self._adj[0][0]
def add_node(self, parent, node):
self._adj[parent].append(node)
def get_children(self, parent):
return self._adj[parent]
#lines = open("input.txt", 'r').readlines()
lines = stdin.readlines()
n = int(lines[0])
def print_found_nodes(found_nodes):
print(" ".join(map(str, found_nodes)))
tree = RootedTree(n)
weights = [0] * (n + 1)
for node, line in enumerate(lines[1:]):
parent, weight = list(map(int, line.split()))
tree.add_node(parent, node + 1)
weights[node + 1] = weight
sum_weights = [0] * (n + 1)
def compute_subtree_sum(node):
global tree
global sum_weights
weight_sum = weights[node]
for child in tree.get_children(node):
compute_subtree_sum(child)
weight_sum = weight_sum + sum_weights[child]
sum_weights[node] = weight_sum
compute_subtree_sum(tree.root)
def divisible_by_three(value):
return abs(value) % 3 == 0
total_sum = sum_weights[tree.root]
if n == 12873:
print(str(sum_weights[3656]) + " " + str(sum_weights[1798]))
elif not divisible_by_three(total_sum):
print("-1")
else:
part_sum = total_sum / 3
found_nodes = []
def find_thirds(node):
global sum_weights
global tree
global total_sum
global found_nodes
if len(found_nodes) == 2:
return
for child in tree.get_children(node):
if sum_weights[child] == total_sum / 3:
found_nodes.append(child)
if len(found_nodes) == 2:
return
else:
find_thirds(child)
found_nodes = [tree.root]
while len(found_nodes) == 1:
node = found_nodes[0]
found_nodes = []
find_thirds(node)
if len(found_nodes) == 2:
#if n == 12873:
# print("a " + str(sum_weights[tree.root]) + " " + str(sum_weights[found_nodes[0]]) + " " + str(sum_weights[found_nodes[1]]))
print_found_nodes(found_nodes)
else:
def find_one_third(node):
global sum_weights
global tree
global total_sum
global found_nodes
if len(found_nodes) >= 2:
return
for child in tree.get_children(node):
if sum_weights[child] == total_sum / 3:
found_nodes.append(child)
return
find_one_third(child)
def find_two_thirds(node):
global sum_weights
global tree
global total_sum
global found_nodes
if len(found_nodes) == 2:
return
for child in tree.get_children(node):
if sum_weights[child] == (total_sum * 2) / 3:
found_nodes.append(child)
find_one_third(child)
if len(found_nodes) == 2:
return
else:
found_nodes = []
find_two_thirds(child)
found_nodes = [tree.root]
while len(found_nodes) == 1:
node = found_nodes[0]
found_nodes = []
find_two_thirds(node)
if len(found_nodes) == 2:
#if n == 12873:
# print("b " + str(sum_weights[tree.root]) + " " + str(sum_weights[found_nodes[0]]) + " " + str(sum_weights[found_nodes[1]]))
print_found_nodes(found_nodes)
else:
print("-1")
``` | instruction | 0 | 99,957 | 7 | 199,914 |
No | output | 1 | 99,957 | 7 | 199,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once at New Year Dima had a dream in which he was presented a fairy garland. A garland is a set of lamps, some pairs of which are connected by wires. Dima remembered that each two lamps in the garland were connected directly or indirectly via some wires. Furthermore, the number of wires was exactly one less than the number of lamps.
There was something unusual about the garland. Each lamp had its own brightness which depended on the temperature of the lamp. Temperatures could be positive, negative or zero. Dima has two friends, so he decided to share the garland with them. He wants to cut two different wires so that the garland breaks up into three parts. Each part of the garland should shine equally, i. e. the sums of lamps' temperatures should be equal in each of the parts. Of course, each of the parts should be non-empty, i. e. each part should contain at least one lamp.
<image>
Help Dima to find a suitable way to cut the garland, or determine that this is impossible.
While examining the garland, Dima lifted it up holding by one of the lamps. Thus, each of the lamps, except the one he is holding by, is now hanging on some wire. So, you should print two lamp ids as the answer which denote that Dima should cut the wires these lamps are hanging on. Of course, the lamp Dima is holding the garland by can't be included in the answer.
Input
The first line contains single integer n (3 ≤ n ≤ 106) — the number of lamps in the garland.
Then n lines follow. The i-th of them contain the information about the i-th lamp: the number lamp ai, it is hanging on (and 0, if is there is no such lamp), and its temperature ti ( - 100 ≤ ti ≤ 100). The lamps are numbered from 1 to n.
Output
If there is no solution, print -1.
Otherwise print two integers — the indexes of the lamps which mean Dima should cut the wires they are hanging on. If there are multiple answers, print any of them.
Examples
Input
6
2 4
0 5
4 2
2 1
1 1
4 2
Output
1 4
Input
6
2 4
0 6
4 2
2 1
1 1
4 2
Output
-1
Note
The garland and cuts scheme for the first example:
<image>
Submitted Solution:
```
def ri():
return map(int, input().split())
class Node:
def __init__(self):
self.adj = set()
self.t = 0
self.v = 0
def dfs(node, s):
global cut
if cut != -1:
return
node[s].v = 1
for a in node[s].adj:
if node[a].v == 0:
node[a].v = 1
dfs(node, a)
if cut != -1:
return
node[s].t += node[a].t
if node[s].t == val:
cut = s
n = int(input())
node = []
par = []
for i in range(0, n+1):
node.append(Node())
val = 0
par = [0]*(n+1)
for i in range(1, n+1):
p, t = ri()
if p == 0:
root = i
node[i].t = t
val += t
if p != 0:
par[i] = p
node[i].adj.add(p)
node[p].adj.add(i)
cut = -1
if val%3:
print(-1)
exit()
else:
val = val//3
dfs(node, root)
ans = []
ans.append(cut)
for i in range(1, n+1):
node[i].s = 0
node[i].v = 0
node[par[cut]].adj.remove(cut)
cut = -1
dfs(node, root)
ans.append(cut)
print(' '.join(map(str, ans)))
``` | instruction | 0 | 99,958 | 7 | 199,916 |
No | output | 1 | 99,958 | 7 | 199,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once at New Year Dima had a dream in which he was presented a fairy garland. A garland is a set of lamps, some pairs of which are connected by wires. Dima remembered that each two lamps in the garland were connected directly or indirectly via some wires. Furthermore, the number of wires was exactly one less than the number of lamps.
There was something unusual about the garland. Each lamp had its own brightness which depended on the temperature of the lamp. Temperatures could be positive, negative or zero. Dima has two friends, so he decided to share the garland with them. He wants to cut two different wires so that the garland breaks up into three parts. Each part of the garland should shine equally, i. e. the sums of lamps' temperatures should be equal in each of the parts. Of course, each of the parts should be non-empty, i. e. each part should contain at least one lamp.
<image>
Help Dima to find a suitable way to cut the garland, or determine that this is impossible.
While examining the garland, Dima lifted it up holding by one of the lamps. Thus, each of the lamps, except the one he is holding by, is now hanging on some wire. So, you should print two lamp ids as the answer which denote that Dima should cut the wires these lamps are hanging on. Of course, the lamp Dima is holding the garland by can't be included in the answer.
Input
The first line contains single integer n (3 ≤ n ≤ 106) — the number of lamps in the garland.
Then n lines follow. The i-th of them contain the information about the i-th lamp: the number lamp ai, it is hanging on (and 0, if is there is no such lamp), and its temperature ti ( - 100 ≤ ti ≤ 100). The lamps are numbered from 1 to n.
Output
If there is no solution, print -1.
Otherwise print two integers — the indexes of the lamps which mean Dima should cut the wires they are hanging on. If there are multiple answers, print any of them.
Examples
Input
6
2 4
0 5
4 2
2 1
1 1
4 2
Output
1 4
Input
6
2 4
0 6
4 2
2 1
1 1
4 2
Output
-1
Note
The garland and cuts scheme for the first example:
<image>
Submitted Solution:
```
import sys
n = int(input())
temp = [-1] * n
parent = [-1] * n
treeTemp = [-1] * n
childStart = [-1] * n
childNext = [-1] * n
sumTemp = 0
for i in range(n):
parent[i], temp[i] = (int(x) for x in input().split())
treeTemp[i] = temp[i]
parent[i] -= 1
sumTemp += temp[i]
if (parent[i] >= 0):
if (childStart[parent[i]] < 0):
childStart[parent[i]] = i
else:
childNext[i] = childStart[parent[i]]
childStart[parent[i]] = i
if(parent[i] == -1):
root = i
if (sumTemp % 3 != 0):
print(-1)
sys.exit()
else:
iterator = root
while (iterator != -1):
if (childStart[iterator] == -1):
if (iterator != root):
treeTemp[parent[iterator]] += treeTemp[iterator]
iterator = parent[iterator]
else:
iterator = childStart[iterator]
childStart[parent[iterator]] = childNext[childStart[parent[iterator]]]
treesCount = 0
treeInd = 0
sumTemp /= 3
for i in range(n):
if (treeTemp[i] == sumTemp):
if (treesCount == 0):
treesCount = 1
treeInd = i
else:
j = i
while (parent[j] != treeInd and parent[j] != -1):
j = parent[j]
if (j == root):
print(treeInd + 1, i + 1)
sys.exit()
print(-1)
``` | instruction | 0 | 99,959 | 7 | 199,918 |
No | output | 1 | 99,959 | 7 | 199,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once at New Year Dima had a dream in which he was presented a fairy garland. A garland is a set of lamps, some pairs of which are connected by wires. Dima remembered that each two lamps in the garland were connected directly or indirectly via some wires. Furthermore, the number of wires was exactly one less than the number of lamps.
There was something unusual about the garland. Each lamp had its own brightness which depended on the temperature of the lamp. Temperatures could be positive, negative or zero. Dima has two friends, so he decided to share the garland with them. He wants to cut two different wires so that the garland breaks up into three parts. Each part of the garland should shine equally, i. e. the sums of lamps' temperatures should be equal in each of the parts. Of course, each of the parts should be non-empty, i. e. each part should contain at least one lamp.
<image>
Help Dima to find a suitable way to cut the garland, or determine that this is impossible.
While examining the garland, Dima lifted it up holding by one of the lamps. Thus, each of the lamps, except the one he is holding by, is now hanging on some wire. So, you should print two lamp ids as the answer which denote that Dima should cut the wires these lamps are hanging on. Of course, the lamp Dima is holding the garland by can't be included in the answer.
Input
The first line contains single integer n (3 ≤ n ≤ 106) — the number of lamps in the garland.
Then n lines follow. The i-th of them contain the information about the i-th lamp: the number lamp ai, it is hanging on (and 0, if is there is no such lamp), and its temperature ti ( - 100 ≤ ti ≤ 100). The lamps are numbered from 1 to n.
Output
If there is no solution, print -1.
Otherwise print two integers — the indexes of the lamps which mean Dima should cut the wires they are hanging on. If there are multiple answers, print any of them.
Examples
Input
6
2 4
0 5
4 2
2 1
1 1
4 2
Output
1 4
Input
6
2 4
0 6
4 2
2 1
1 1
4 2
Output
-1
Note
The garland and cuts scheme for the first example:
<image>
Submitted Solution:
```
def tot_brightness(cur):
if visited[cur]:
return 0
visited[cur] = 1
if cur+1 in parent:
nex = parent.index(cur+1)
return brightness[nex] + tot_brightness(nex)
else:
return 0
def getRightChild(cur, tsum):
if tsum == req_sum:
return lchild(cur)
elif tsum > req_sum:
return -1
else:
if cur+1 in parent:
nex = parent.index(cur+1)
return getRightChild(nex, tsum+brightness[nex])
def lchild(cur):
if cur+1 in parent:
return parent.index(cur+1)
else:
return -1
def child(cur):
pos = 0
while cur+1 in parent[pos:]:
k = parent[pos:].index(cur+1)
if visited[pos+k] == 0:
return pos+k
pos += k+1
return -1
def getLeftChild(cur, tsum):
if (right_sum - tsum) == req_sum:
return child(cur), tsum
elif (right_sum - tsum) < req_sum:
return -1, 0
else:
visited[cur] = 1
nex = 0
while cur+1 in parent[nex:]:
k = parent[nex:].index(cur+1)
print('>>>>', nex, k)
if visited[nex+k] == 0:
return getLeftChild(nex+k, tsum+brightness[nex+k])
nex += k+1
n = int(input())
parent = []
brightness = []
for i in range(n):
a, b = input().split()
parent.append(int(a))
brightness.append(int(b))
visited = [0]*len(parent)
start = parent.index(0)
left_sum = tot_brightness(start)
right_sum = 0
for i in range(len(visited)):
if visited[i] == 0:
right_sum += brightness[i]
tot_sum = left_sum + right_sum + brightness[start]
if tot_sum%3 == 0:
req_sum = tot_sum / 3
mid_sum = 0
visited[start] = 0
left, midsum = getLeftChild(start, mid_sum)
if left != -1:
right = getRightChild(start, mid_sum+brightness[start])
if right != 1:
print(right+1, left+1)
else:
print(-1)
else:
print(-1)
else:
print(-1)
``` | instruction | 0 | 99,960 | 7 | 199,920 |
No | output | 1 | 99,960 | 7 | 199,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,045 | 7 | 200,090 |
Tags: greedy, implementation
Correct Solution:
```
import sys
n,m = [int(x) for x in input().split(' ')]
a=[]
for i in range(n):
b=[]
s = input().strip()
for j in range(m):
if s[j]=='#':
b.append(j)
if b not in a:
a.append(b)
c=[0]*m
ans=True
#print(a)
for i in a:
for j in i:
c[j]+=1
if c[j]>1:
ans=False
break
if ans:
print("Yes")
else:
print("No")
``` | output | 1 | 100,045 | 7 | 200,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,046 | 7 | 200,092 |
Tags: greedy, implementation
Correct Solution:
```
par = input()
par = list(map(int, par.split()))
n, m = par[0], par[1]
mat = []
for i in range(n):
row = list(input())
mat.append(row)
def posColor(n, m, mt):
for i in range(n):
for j in range(m):
if mt[i][j] == '#':
for x in range(i + 1, n):
if mt[x][j] == '#':
for y in range(m):
if mt[i][y] != mt[x][y]:
return 'No'
return 'Yes'
print(posColor(n, m, mat))
``` | output | 1 | 100,046 | 7 | 200,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,047 | 7 | 200,094 |
Tags: greedy, implementation
Correct Solution:
```
from collections import defaultdict
r,c=map(int,input().split())
h=defaultdict(list)
flg=[]
for i in range(r):
st=input()
st=str(st)
y=[]
for j in range(c):
if st[j]=='#':
y.append(j)
if y in list(h.values()):
flg.append(1)
# print(1)
else:
for k in y:
for item in list(h.values()):
if k in item:
flg.append(0)
# print(2)
break
h[i]=y
# print(list(h.values()),y)
if 0 in flg:
print('No')
else:
print('Yes')
``` | output | 1 | 100,047 | 7 | 200,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,048 | 7 | 200,096 |
Tags: greedy, implementation
Correct Solution:
```
N, M = map(int, input().split())
grid = []
for _ in range(N):
grid.append(input())
S = [set() for _ in range(N)]
for i in range(N):
for j in range(M):
if grid[i][j] == "#":
S[i].add(j)
solved = [False for _ in range(N)]
used = [False for _ in range(M)]
for i in range(N):
if not solved[i]:
for c in S[i]:
if used[c]:
print("No")
quit()
used[c] = True
for j in range(N):
if S[i] == S[j]:
solved[j] = True
print("Yes")
``` | output | 1 | 100,048 | 7 | 200,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,049 | 7 | 200,098 |
Tags: greedy, implementation
Correct Solution:
```
used =[]
a = []
def dfs(v, c):
vert = v + 100*c
if used[vert] == 0:
used[vert] = 1
a[c].append(v)
for i in range(0, len(g[vert])):
if used[g[vert][i]] == 0:
dfs(g[vert][i] - (1-c)*100, 1-c)
n,m = map(int, input().split())
r = []
for i in range(0,n):
s = input()
r.append(s)
g = []
for i in range (0,200):
g.append([])
used.append(0)
for i in range (0,n):
for j in range (0,m):
if(r[i][j] == "#"):
g[i].append(100 + j)
g[100+j].append(i)
#print(g)
for i in range(0,n):
a = [[],[]]
dfs(i,0)
#print(a)
for j in range (0, len(a[0])):
for k in range (0, len(a[1])):
if(r[a[0][j]][a[1][k]] != "#"):
print ("NO")
raise SystemExit
print("YES")
``` | output | 1 | 100,049 | 7 | 200,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,050 | 7 | 200,100 |
Tags: greedy, implementation
Correct Solution:
```
n, m = map(int, input().split())
x = []
for i in range(n):
q = []
y = input()
for j in y:
q.append(j)
x.append(q)
ind = True
for i in range(n):
for j in range(m):
if x[i][j] == "#":
for k in range(i + 1, n):
if x[k][j] == "#" and x[k] != x[i]:
ind = False
break
if ind:
print("Yes")
else:
print("No")
``` | output | 1 | 100,050 | 7 | 200,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,051 | 7 | 200,102 |
Tags: greedy, implementation
Correct Solution:
```
def partial_match(r1, r2):
n = len(r1)
match_found = False
mismatch_found = False
for i in range(n):
if r1[i] == '#' and r2[i] == '#':
match_found = True
if r1[i] == '#' and r2[i] == '.':
mismatch_found = True
if r2[i] == '#' and r1[i] == '.':
mismatch_found = True
return match_found and mismatch_found
n , m = tuple(map(int, input().split()))
l = []
for i in range(n):
l.append(input().strip())
invalid = False
for i in range(n):
for j in range(n):
if partial_match(l[i], l[j]):
invalid = True
break
if invalid:
print("No")
else:
print("Yes")
``` | output | 1 | 100,051 | 7 | 200,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. | instruction | 0 | 100,052 | 7 | 200,104 |
Tags: greedy, implementation
Correct Solution:
```
def func(ss):
occured = set()
for pr, s in zip(['']+ss[:-1], ss):
if pr == s:
continue
for i, el in enumerate(s):
if el == "#":
if i in occured:
return False
occured.add(i)
return True
n, m = [int(x) for x in input().split()]
ss = [input() for _ in range(n)]
ss.sort()
print("Yes" if func(ss) else "No")
``` | output | 1 | 100,052 | 7 | 200,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
n, m = map(int, input().split())
st = []
for i in range(n):
s = input()
x = set()
for j in range(m):
if s[j] == '#':
x.add(j)
for i in st:
if i & x and i != x:
print('No')
exit()
st.append(x)
print('Yes')
``` | instruction | 0 | 100,053 | 7 | 200,106 |
Yes | output | 1 | 100,053 | 7 | 200,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
n, m = map( int, input().split() )
M = []
for i in range(n):
M.append(input())
Adj = [[] for i in range(n+m)]
for i in range(n):
for j in range(m):
if M[i][j] == '#':
Adj[i].append(n+j)
Adj[n+j].append(i)
p = [-1]*(n+m)
comp = []
answer = 'Yes'
for i in range(n+m):
if p[i] != -1:
continue
L = [i]
comp.append([])
numEdges = 0
numA = 0
numB = 0
while L:
v = L.pop()
comp[-1].append((v, 'A') if v < n else (v-n, 'B'))
if v < n:
numA += 1
else:
numB += 1
for u in Adj[v]:
numEdges += 1
if p[u] == -1 and u != i:
p[u] = v
L.append(u)
#print(comp[-1], numA, numB, numEdges/2)
if numEdges//2 != numA*numB:
answer = 'No'
print(answer)
``` | instruction | 0 | 100,054 | 7 | 200,108 |
Yes | output | 1 | 100,054 | 7 | 200,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
n,m = map(int,input().split())
a = []
f = 0
for i in range(n):a.append(input())
for i in range(n):
for j in range(i+1,n):
r1 = a[i]
r2 = a[j]
f1 = f2 = 0
for k in range(m):
if r1[k] != r2[k]:f1 = 1
if (r1[k] == '#') and (r2[k]=='#'):f2 = 1
if f1 == 1:
if f2 == 1:
f = 1
if f:print('No')
else:print('Yes')
``` | instruction | 0 | 100,055 | 7 | 200,110 |
Yes | output | 1 | 100,055 | 7 | 200,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
n,m=map(int,input().split())
a=[]
b=[]
for i in range(n):
a.append(input())
s=''
for j in a[i]:
if j=='#':
s+='1'
else:
s+='0'
b.append(s)
for i in range(n):
for j in range(i,n):
if a[i]!=a[j]:
if '2' in str(int(b[i]) + int(b[j])):
print('No')
exit()
print('Yes')
``` | instruction | 0 | 100,056 | 7 | 200,112 |
Yes | output | 1 | 100,056 | 7 | 200,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
def table(n, m, lst):
for i in range(n):
for j in range(n):
if lst[i] != lst[j]:
for k in range(m):
if lst[i][z] == lst[j][z] == "#":
return "No"
return "Yes"
n, m = [int(i) for i in input().split()]
b = list()
for i in range(n):
z = list(input())
b.append(z)
``` | instruction | 0 | 100,057 | 7 | 200,114 |
No | output | 1 | 100,057 | 7 | 200,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
from collections import defaultdict as dd, deque
r,c = map(int,input().split())
A = [list(input()) for i in range(r)]
At = [[A[i][j] for i in range(r)] for j in range(c)]
R=r
C=c
def find():
for i in range(R):
for j in range(C):
if A[i][j] == "#":
return i,j
return ()
while True:
pos = find()
if pos:
i,j = pos
rows = [i]
cols = [j]
for x in range(c):
if A[i][x] == '#':
cols.append(x)
for x in range(r):
if A[x][j] == '#':
rows.append(x)
fail = False
row = A[rows[0]]
for x in range(1,len(rows)):
if row != A[rows[x]]:
fail = True
col = At[cols[0]]
for x in range(1,len(cols)):
if col != At[cols[x]]:
fail = True
if fail:
print('No')
break
#print('fill')
for r in rows:
for c in cols:
A[r][c] = '.'
At[c][r] = '.'
#print(cols)
#print(rows)
#for a in A:
# print(''.join(a))
#for a in At:
# print(''.join(a))
else:
print('Yes')
break
``` | instruction | 0 | 100,058 | 7 | 200,116 |
No | output | 1 | 100,058 | 7 | 200,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
n, m = map(int, input().split())
array = []
state = True
for i in range(n):
string = input()
if string.find('##')!=-1:
state = False
if state==True:
print('Yes')
else:
print('No')
``` | instruction | 0 | 100,059 | 7 | 200,118 |
No | output | 1 | 100,059 | 7 | 200,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a rectangular grid of n rows of m initially-white cells each.
Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black.
There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set.
You are to determine whether a valid sequence of operations exists that produces a given final grid.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the number of rows and columns of the grid, respectively.
Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup.
Output
If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5 8
.#.#..#.
.....#..
.#.#..#.
#.#....#
.....#..
Output
Yes
Input
5 5
..#..
..#..
#####
..#..
..#..
Output
No
Input
5 9
........#
#........
..##.#...
.......#.
....#.#.#
Output
No
Note
For the first example, the desired setup can be produced by 3 operations, as is shown below.
<image>
For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column.
Submitted Solution:
```
n, m = map(int, input().split())
l = [input() for i in range(n)]
flag = True
for i in range(n):
if not flag:
break
for j in range(i + 1, n):
if l[i] != l[j]:
for q in range(m):
if l[j][q] == l[i][q] == '#':
print('NO')
flag = False
break
else:
print('YES')
``` | instruction | 0 | 100,060 | 7 | 200,120 |
No | output | 1 | 100,060 | 7 | 200,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.
Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109 + 7).
Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.
Input
The first line of input contains two space-separated integers n and s (1 ≤ n ≤ 20, 0 ≤ s ≤ 1014).
The second line contains n space-separated integers f1, f2, ... fn (0 ≤ fi ≤ 1012).
Output
Output a single integer — the number of ways in which Devu can select the flowers modulo (109 + 7).
Examples
Input
2 3
1 3
Output
2
Input
2 4
2 2
Output
1
Input
3 5
1 3 2
Output
3
Note
Sample 1. There are two ways of selecting 3 flowers: {1, 2} and {0, 3}.
Sample 2. There is only one way of selecting 4 flowers: {2, 2}.
Sample 3. There are three ways of selecting 5 flowers: {1, 2, 2}, {0, 3, 2}, and {1, 3, 1}.
Submitted Solution:
```
MOD = 10**9 +7
def C(n,r):
if n < r:
return 0
n%=MOD
r = min(r,n-r)
ans =1
for i in range(r):
ans = ans*(n-i) % MOD
ans = ans*inv[i+1] % MOD
return ans
n, s = [int(k) for k in input().split()]
f = [int(k) for k in input().split()]
f.insert(0,0)
inv = [0]*50
for i in range(50):
inv[i] = pow(i, MOD - 2, MOD)
ways = 0
for i in range((1<<n)):
#print(bin(i))
sum1=0
sign = 1
for j in range(n):
if (1<<j) &i:
sign *=-1
sum1 +=f[j+1]+1
#print(sum1)
if sum1 <= s:
ways+=sign*C(n+(s-sum1)-1,s-sum1) % MOD
#print(ways)
print(ways % MOD)
``` | instruction | 0 | 100,695 | 7 | 201,390 |
No | output | 1 | 100,695 | 7 | 201,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Devu wants to decorate his garden with flowers. He has purchased n boxes, where the i-th box contains fi flowers. All flowers in a single box are of the same color (hence they are indistinguishable). Also, no two boxes have flowers of the same color.
Now Devu wants to select exactly s flowers from the boxes to decorate his garden. Devu would like to know, in how many different ways can he select the flowers from each box? Since this number may be very large, he asks you to find the number modulo (109 + 7).
Devu considers two ways different if there is at least one box from which different number of flowers are selected in these two ways.
Input
The first line of input contains two space-separated integers n and s (1 ≤ n ≤ 20, 0 ≤ s ≤ 1014).
The second line contains n space-separated integers f1, f2, ... fn (0 ≤ fi ≤ 1012).
Output
Output a single integer — the number of ways in which Devu can select the flowers modulo (109 + 7).
Examples
Input
2 3
1 3
Output
2
Input
2 4
2 2
Output
1
Input
3 5
1 3 2
Output
3
Note
Sample 1. There are two ways of selecting 3 flowers: {1, 2} and {0, 3}.
Sample 2. There is only one way of selecting 4 flowers: {2, 2}.
Sample 3. There are three ways of selecting 5 flowers: {1, 2, 2}, {0, 3, 2}, and {1, 3, 1}.
Submitted Solution:
```
#!/usr/bin/env python3
import math
import operator
from functools import reduce
def prod(iterable):
return reduce(operator.mul, iterable, 1)
def c(n, k):
return math.factorial(n) / math.factorial(k) * math.factorial(n - k)
def solve(s, ns):
n = sum(ns)
return c(n, s)
def main():
n, s = list(map(int, input().split()))
ns = list(map(int, input().split()))
answer = solve(s, ns) / (n - 1)
print(int(answer) % 1000000007)
if __name__ == '__main__':
main()
``` | instruction | 0 | 100,696 | 7 | 201,392 |
No | output | 1 | 100,696 | 7 | 201,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
BerPhone X is almost ready for release with n applications being preinstalled on the phone. A category of an application characterizes a genre or a theme of this application (like "game", "business", or "education"). The categories are given as integers between 1 and n, inclusive; the i-th application has category c_i.
You can choose m — the number of screens and s — the size of each screen. You need to fit all n icons of the applications (one icon representing one application) meeting the following requirements:
* On each screen, all the icons must belong to applications of the same category (but different screens can contain icons of applications of the same category);
* Each screen must be either completely filled with icons (the number of icons on the screen is equal to s) or almost filled with icons (the number of icons is equal to s-1).
Your task is to find the minimal possible number of screens m.
Input
The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains an integer n (1 ≤ n ≤ 2⋅10^6) — the number of the icons. The second line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n), where c_i is the category of the i-th application.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 2⋅10^6.
Output
Print t integers — the answers to the given test cases in the order they follow in the input. The answer to a test case is an integer m — the minimum number of screens on which all n icons can be placed satisfying the given requirements.
Example
Input
3
11
1 5 1 5 1 5 1 1 1 1 5
6
1 2 2 2 2 1
5
4 3 3 1 2
Output
3
3
4
Note
In the first test case of the example, all the icons can be placed on three screens of size 4: a screen with 4 icons of the category 1, a screen with 3 icons of the category 1, and a screen with 4 icons of the category 5.
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
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")
# ------------------- fast io --------------------
from collections import defaultdict
for j in range(int(input())):
n=int(input());vals=list(map(int,input().split()))
dict1=defaultdict(int)
for s in range(n):
dict1[vals[s]]+=1
minnum=min(dict1.values())+1;minscreens=2*10**6 +1
for s in range(1,minnum+1):
screens=0;found=True
for i in dict1.values():
mult_1=(-i)%s
mult=(i-mult_1*(s-1))//s
if mult<0:
found=False;break
else:
screens+=mult_1+mult
if found==True:
minscreens=min(minscreens,screens)
print(minscreens)
``` | instruction | 0 | 101,326 | 7 | 202,652 |
Yes | output | 1 | 101,326 | 7 | 202,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
BerPhone X is almost ready for release with n applications being preinstalled on the phone. A category of an application characterizes a genre or a theme of this application (like "game", "business", or "education"). The categories are given as integers between 1 and n, inclusive; the i-th application has category c_i.
You can choose m — the number of screens and s — the size of each screen. You need to fit all n icons of the applications (one icon representing one application) meeting the following requirements:
* On each screen, all the icons must belong to applications of the same category (but different screens can contain icons of applications of the same category);
* Each screen must be either completely filled with icons (the number of icons on the screen is equal to s) or almost filled with icons (the number of icons is equal to s-1).
Your task is to find the minimal possible number of screens m.
Input
The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains an integer n (1 ≤ n ≤ 2⋅10^6) — the number of the icons. The second line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n), where c_i is the category of the i-th application.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 2⋅10^6.
Output
Print t integers — the answers to the given test cases in the order they follow in the input. The answer to a test case is an integer m — the minimum number of screens on which all n icons can be placed satisfying the given requirements.
Example
Input
3
11
1 5 1 5 1 5 1 1 1 1 5
6
1 2 2 2 2 1
5
4 3 3 1 2
Output
3
3
4
Note
In the first test case of the example, all the icons can be placed on three screens of size 4: a screen with 4 icons of the category 1, a screen with 3 icons of the category 1, and a screen with 4 icons of the category 5.
Submitted Solution:
```
from collections import defaultdict,Counter
from math import inf,ceil
for _ in range(int(input())):
n=int(input())
ci=list(map(int,input().split()))
freq=Counter(ci)
freqlist=list(freq.values())
minimum=inf
for i in freqlist:
minimum=min(minimum,i)
for screen in range(minimum+1,0,-1):
flag=1
ans=0
for ii in freqlist:
nos=ceil(ii/screen)
val1=(screen-1)*nos
val2=nos*screen
if ii<=val2 and ii>=val1:
ans+=nos
else:
flag=0
ans=0
break
if flag:
print(ans)
break
``` | instruction | 0 | 101,327 | 7 | 202,654 |
Yes | output | 1 | 101,327 | 7 | 202,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
BerPhone X is almost ready for release with n applications being preinstalled on the phone. A category of an application characterizes a genre or a theme of this application (like "game", "business", or "education"). The categories are given as integers between 1 and n, inclusive; the i-th application has category c_i.
You can choose m — the number of screens and s — the size of each screen. You need to fit all n icons of the applications (one icon representing one application) meeting the following requirements:
* On each screen, all the icons must belong to applications of the same category (but different screens can contain icons of applications of the same category);
* Each screen must be either completely filled with icons (the number of icons on the screen is equal to s) or almost filled with icons (the number of icons is equal to s-1).
Your task is to find the minimal possible number of screens m.
Input
The first line contains an integer t (1 ≤ t ≤ 10 000) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains an integer n (1 ≤ n ≤ 2⋅10^6) — the number of the icons. The second line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n), where c_i is the category of the i-th application.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 2⋅10^6.
Output
Print t integers — the answers to the given test cases in the order they follow in the input. The answer to a test case is an integer m — the minimum number of screens on which all n icons can be placed satisfying the given requirements.
Example
Input
3
11
1 5 1 5 1 5 1 1 1 1 5
6
1 2 2 2 2 1
5
4 3 3 1 2
Output
3
3
4
Note
In the first test case of the example, all the icons can be placed on three screens of size 4: a screen with 4 icons of the category 1, a screen with 3 icons of the category 1, and a screen with 4 icons of the category 5.
Submitted Solution:
```
import sys
import math
from collections import defaultdict
def get_factors(num):
a=set()
for i in range(1,int(math.sqrt(num))+2):
if num%i==0:
a.add(i)
a.add(num//i)
return a
t=int(sys.stdin.readline())
for _ in range(t):
n=int(sys.stdin.readline())
dic=defaultdict(int)
arr=list(map(int,sys.stdin.readline().split()))
for i in range(n):
dic[arr[i]]+=1
'''factors=[]
#print(dic,'dic')
for i in dic:
a=get_factors(dic[i])
b=get_factors(dic[i]+1)
a=a.union(b)
factors.append(a)
m=len(factors)
#print(factors,'factors')
for i in range(1,m):
factors[0]=factors[0].intersection(factors[i])
maxsize=max(factors[0])'''
l=[]
for i in dic:
l.append(dic[i])
l.sort()
x=l[0]
ans=sum(l)
m=len(l)
#print(l,'l')
for i in range(1,x+2):
s,z=0,True
for j in range(m):
a=math.ceil(l[j]/i)
if l[j]<(i-1)*a:
z=False
break
else:
s+=a
if z:
ans=min(ans,s)
print(int(ans))
``` | instruction | 0 | 101,328 | 7 | 202,656 |
Yes | output | 1 | 101,328 | 7 | 202,657 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.