message stringlengths 2 23.8k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa | instruction | 0 | 71,808 | 0 | 143,616 |
Tags: binary search, brute force, combinatorics, implementation, math
Correct Solution:
```
arr=[1]
zz=10**5+100
for i in range(1,zz):
arr.append(arr[-1]+i)
from bisect import bisect_left as bl, bisect_right as br
for _ in range(int(input())):
n,k=map(int,input().split())
ind=bl(arr,k)
if arr[ind]!=k:
ind-=1
if arr[ind+1]==k:
ans=0
else: ans=k-arr[ind]
x=n-1-abs(ans)
y=n-ind-2
rans=['a']*n
rans[x]='b'
rans[y]='b'
print(''.join(rans))
``` | output | 1 | 71,808 | 0 | 143,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa | instruction | 0 | 71,809 | 0 | 143,618 |
Tags: binary search, brute force, combinatorics, implementation, math
Correct Solution:
```
t = int(input())
for i in range(t):
N,K=[int(i) for i in input().split(" ")]
a=int(((1+8*K)**(0.5))/2 - 0.5)
b=K-a*(a+1)//2
if b==0:
print("a"*(N-a-1)+"bb"+"a"*(a+1-2))
else:
print("a"*(N-a-2)+"b"+"a"*(a-b+1)+"b"+"a"*(b-1))
``` | output | 1 | 71,809 | 0 | 143,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa | instruction | 0 | 71,810 | 0 | 143,620 |
Tags: binary search, brute force, combinatorics, implementation, math
Correct Solution:
```
from math import floor, sqrt
def sigma(n):
return n * (n - 1) // 2
def sigma_inv(n):
f = 1 / 2 + sqrt(2 * n - 2 + 1 / 4)
return floor(f)
n = int(input())
for _ in range(n):
s, k = map(int, input().split())
ans = ["a"] * s
b2 = k - sigma(sigma_inv(k))
b1 = floor(sigma_inv(k)) + 1
ans[-b2] = 'b'
ans[-b1] = 'b'
print(*ans, sep='')
``` | output | 1 | 71,810 | 0 | 143,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa | instruction | 0 | 71,811 | 0 | 143,622 |
Tags: binary search, brute force, combinatorics, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
x = n * (n - 1) // 2
c = n-1
while x - c >= k:
x -= c
c -= 1
l = ['a'] * n
l[c] = 'b'
l[c+(k-x-1)] = 'b'
print(*l[::-1], sep='')
``` | output | 1 | 71,811 | 0 | 143,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa | instruction | 0 | 71,812 | 0 | 143,624 |
Tags: binary search, brute force, combinatorics, implementation, math
Correct Solution:
```
from math import floor, sqrt
t = int(input())
ans = []
def find_first(k):
for i in range(k+2):
if i*(i-1)/2>=k:
return i
def find_second(k, first):
return int(k-(first-1)*(first-2)/2)
for _ in range(t):
n, k = map(int, input().split())
a = ['a' for i in range(n)]
first = find_first(k)
second = find_second(k, first)
a[first-1] = a[second-1] = 'b'
ans.append(''.join(reversed(a)))
print(*ans, sep = '\n')
``` | output | 1 | 71,812 | 0 | 143,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
arr=[]
arr.append(2)
len=1
while(len<=1e5):
arr.append((len*(len+1)/2))
len+=1
def binarySearchCount(key):
left = 0
right = len
mid = 0
while (left < right):
mid = (right + left)//2
if (arr[mid] == key):
while (mid + 1<len and arr[mid + 1] == key):
mid+= 1
break
elif (arr[mid] > key):
right = mid
else:
left = mid + 1
while (mid > -1 and arr[mid] > key):
mid-= 1
return mid
t=int(input())
while(t>0):
n,k=map(int, input().split())
temp=binarySearchCount(k)
if int(k-arr[temp])==0:
temp2=temp-1
else:
temp2=int(k-arr[temp])-1
temp+=1
out=[]
for i in range(n):
if i==temp or i==temp2:
out.append('b')
else:
out.append('a')
for i in range(n):
print(out[n-1-i],end="")
print()
t-=1
``` | instruction | 0 | 71,813 | 0 | 143,626 |
Yes | output | 1 | 71,813 | 0 | 143,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
# May the SpeedForce be with us
sum=[0]*(100007)
mad=0
for i in range(1,100005):
mad+=i
sum[i]=mad
from bisect import bisect_left
def BinarySearch(a, x):
i = bisect_left(a, x)
if i:
return (i-1)
else:
return -1
for i in range(int(input())):
n,k=map(int,input().split())
l=BinarySearch(sum,k)+1
tup=(l+1,(l)-(sum[l]-k))
#print(tup)
res=['a']*(n)
for i in tup:
res[n-i]='b'
print(''.join(res))
``` | instruction | 0 | 71,814 | 0 | 143,628 |
Yes | output | 1 | 71,814 | 0 | 143,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
n=int(input())
for i in range(n):
a,b=map(int,input().split())
c=1
while((c*(c+1))/2 < b):
c+=1
x=b-(c*(c-1))//2
for j in range(a):
if j==a-x or j==a-c-1:
print('b',end='')
else:
print('a',end='')
print()
``` | instruction | 0 | 71,815 | 0 | 143,630 |
Yes | output | 1 | 71,815 | 0 | 143,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
t = int(input())
while t:
t-=1
n,k = map(int,input().split())
i = n-2
s = ['a']*n
while i>=0:
if k<=(n-i-1):
s[i] = 'b'
s[n - k] = 'b'
break
k -= (n - i - 1)
i -= 1
print("".join(s))
"""
https://codeforces.com/blog/entry/75246?#comment-594464
"""
``` | instruction | 0 | 71,816 | 0 | 143,632 |
Yes | output | 1 | 71,816 | 0 | 143,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
import math
t= int(input())
for i in range(t):
n,k=map(int,input().split())
s=n*"a"
r=int(math.sqrt(2*k))-1
s=s[:r+1]+"b" + s[r+2:]
t=k-(r*(r+1)//2 )
s=s[:t-1]+"b" + s[t:]
print((s[::-1]))
``` | instruction | 0 | 71,817 | 0 | 143,634 |
No | output | 1 | 71,817 | 0 | 143,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
def formation(L):
n = len(L)
i = n - 2
while i >= 0 and L[i] >= L[i + 1]:
i -= 1
if i == -1:
return False
j = i + 1
while j < n and L[j] > L[i]:
j += 1
j -= 1
L[i], L[j] = L[j], L[i]
left = i + 1
right = n - 1
while left < right:
L[left], L[right] = L[right], L[left]
left += 1
right -= 1
return True
def solve(string, n):
string = list(string)
new_string = []
string.sort()
j = 2
while formation(string):
new_string = string
if j == n:
break
j += 1
print(''.join(new_string))
t=int(input())
for _ in range(t):
n,k=map(int, input ().split())
s=""
for i in range(n-2):
s=s+"a"
s=s+"bb"
solve(s,k)
``` | instruction | 0 | 71,818 | 0 | 143,636 |
No | output | 1 | 71,818 | 0 | 143,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
t = int(input())
for e in range(t):
n, k = [int(x) for x in input().split()]
s = ["a"] * n
cont = 0
aux = 0
while True:
if aux > k:
aux -= cont
break
cont += 1
aux += cont
tmp = k - aux
if tmp > 0:
pos1 = n - 1 - cont
pos2 = n - tmp
else:
cont -= 1
pos1 = n - 1 - cont
pos2 = pos1 + 1
s[pos1] = "b"
s[pos2] = "b"
for e in s:
print(e, end=" ")
print()
``` | instruction | 0 | 71,819 | 0 | 143,638 |
No | output | 1 | 71,819 | 0 | 143,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For the given integer n (n > 2) let's write down all the strings of length n which contain n-2 letters 'a' and two letters 'b' in lexicographical (alphabetical) order.
Recall that the string s of length n is lexicographically less than string t of length n, if there exists such i (1 ≤ i ≤ n), that s_i < t_i, and for any j (1 ≤ j < i) s_j = t_j. The lexicographic comparison of strings is implemented by the operator < in modern programming languages.
For example, if n=5 the strings are (the order does matter):
1. aaabb
2. aabab
3. aabba
4. abaab
5. ababa
6. abbaa
7. baaab
8. baaba
9. babaa
10. bbaaa
It is easy to show that such a list of strings will contain exactly (n ⋅ (n-1))/(2) strings.
You are given n (n > 2) and k (1 ≤ k ≤ (n ⋅ (n-1))/(2)). Print the k-th string from the list.
Input
The input contains one or more test cases.
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. Then t test cases follow.
Each test case is written on the the separate line containing two integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ min(2⋅10^9, (n ⋅ (n-1))/(2)).
The sum of values n over all test cases in the test doesn't exceed 10^5.
Output
For each test case print the k-th string from the list of all described above strings of length n. Strings in the list are sorted lexicographically (alphabetically).
Example
Input
7
5 1
5 2
5 8
5 10
3 1
3 2
20 100
Output
aaabb
aabab
baaba
bbaaa
abb
bab
aaaaabaaaaabaaaaaaaa
Submitted Solution:
```
t = int(input())
for x in range(t):
n, k = map(int, input().split())
res = ['a'] * n
s = int((n * (n - 1))/2 )
mark = n-1
mark1 = -1
if s != k:
for i in range(n - 1, -1, -1):
s -= i
mark1 += 1
if s == k:
mark = mark1+1
break
if s < k:
mark1 -= 1
mark = k-s
break
res[mark1+1] = 'b'
res[n-mark] = 'b'
e = ''.join(map(str,res))
print(e)
``` | instruction | 0 | 71,820 | 0 | 143,640 |
No | output | 1 | 71,820 | 0 | 143,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,883 | 0 | 143,766 |
Tags: greedy, implementation
Correct Solution:
```
from sys import stdin ,stdout
from os import path
rd = lambda:stdin.readline().strip()
wr = stdout.write
if(path.exists('input.txt')):
stdin = open("input.txt","r")
import time ,math
#------------------------------------=
from collections import defaultdict
for _ in range(int(rd())):
n,m = map(int,rd().split())
x = rd()
arr = []
out = 1
for i in range(len(x)):
if x[i] == '*' :
arr.append(i)
j = 0
for i in range(1,len(arr)):
if arr[i]-arr[j] < m:
if i == len(arr)-1:
out+=1
continue
elif arr[i]-arr[j] > m:
if (arr[i]- arr[i-1]) == m:
out+=1
j = i
else:
if i == len(arr)-1:
out+=1
j = i-1
out+=1
else:
out+= 1
j = i
print(out)
``` | output | 1 | 71,883 | 0 | 143,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,884 | 0 | 143,768 |
Tags: greedy, implementation
Correct Solution:
```
In=input
lin=lambda : map(int,In().split())
for _ in range(int(In())):
_,k=lin()
m=In()
j=1
i=m.index('*')
while i<m.rfind('*'):
i=m.rfind('*',i,i+k+1)
j+=1
print(j)
``` | output | 1 | 71,884 | 0 | 143,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,885 | 0 | 143,770 |
Tags: greedy, implementation
Correct Solution:
```
from collections import deque
for _ in range(int(input())):
n,k=map(int,input().split())
s=input()
queue=deque()
front=-1
back=-1
for i in range(n-1,-1,-1):
if s[i]=="*":
back=i
break
for i in range(n):
if s[i]=="*":
front=i
break
if front==back:
print(1)
continue
ans=2
dis=0
temp=-1
for i in range(front+1,back,1):
if s[i]=="*":
dis+=1
if dis<k:
temp=i
elif dis==k:
temp=-1
ans+=1
dis=0
else:
dis+=1
if dis==k:
ans+=1
dis=i-temp
temp=-1
ans2 = 2
dis = 0
temp = -1
for i in range(back-1,front,-1):
if s[i]=="*":
dis+=1
if dis<k:
temp=i
elif dis==k:
temp=-1
ans2+=1
dis=0
else:
dis+=1
if dis==k:
ans2+=1
dis = abs(i - temp)
temp=-1
print(min(ans,ans2))
``` | output | 1 | 71,885 | 0 | 143,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,886 | 0 | 143,772 |
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
s=str(input())
p=s.find("*")
q=s.rfind("*")
if p==q:
print(1)
else:
ans=2
while q-p>k:
p=s.rfind("*",0,p+k+1)
ans+=1
print(ans)
``` | output | 1 | 71,886 | 0 | 143,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,887 | 0 | 143,774 |
Tags: greedy, implementation
Correct Solution:
```
Tt=int(input())
for ii in range(Tt):
n,k=map(int,input().split())
s=input()
c=s.count('*')
if c==1:
print(1)
elif c==2:
print(2)
else:
x=s.find('*')
l=s[::-1]
y=l.find('*')
cnt=2
n=n-y
x+=k
while(x<n-1):
if s[x]=='*':
cnt+=1
x+=k
else:x-=1
print(cnt)
``` | output | 1 | 71,887 | 0 | 143,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,888 | 0 | 143,776 |
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
a = []
for i in range(n):
if s[i] == '*':a.append(i)
fi = a[0]
ans=1
a = a[1:]
prev=fi
for i in range(len(a)):
el = a[i]
if el-prev>k:
prev=a[i-1]
ans+=1
if a and fi!=a[-1]:ans+=1
print(ans)
``` | output | 1 | 71,888 | 0 | 143,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,889 | 0 | 143,778 |
Tags: greedy, implementation
Correct Solution:
```
t=int(input())
for i in range(t):
n,k=[int(x) for x in input().split()]
retazec=input()
for i in range(n):
if retazec[i]=='*':
prva=i
break
for i in range(n):
if retazec[n-1-i]=='*':
posledna=n-1-i
break
i=prva
pocet=1
while i!=posledna:
for j in range(1,1+min(k,n-1-i)):
if retazec[i+j]=='*':
teraz=i+j
i=teraz
pocet+=1
print(pocet)
``` | output | 1 | 71,889 | 0 | 143,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1 | instruction | 0 | 71,890 | 0 | 143,780 |
Tags: greedy, implementation
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 4 18:13:28 2021
@author: suneelvarma
"""
def answer(st,k):
ss,es = st.find('*'),st.rfind('*')
count = 1
i = ss
while i < es:
if st[i] == '*': count += 1
nextStar = min(len(st)-1,i+k)
while st[nextStar] != '*': nextStar -= 1
i = nextStar
return count
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n,k = tuple(map(int,input().split()))
st = input()
print(answer(st,k))
``` | output | 1 | 71,890 | 0 | 143,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
for test_case in range(int(input())):
n, k = map(int,input().split())
s=input()
c=1
f=s.find("*")
# print(f,s[0])
for i in range(n):
if f+k<n and s[f+k]=="*":
c+=1
f=f+k
else:
t=f+k
if t>=n:
t=n-1
for j in range(t,f,-1):
if s[j]=="*":
c+=1
f=j
break
print(c)
``` | instruction | 0 | 71,891 | 0 | 143,782 |
Yes | output | 1 | 71,891 | 0 | 143,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
for _ in range(int(input())):
n, k = list(map(int, input().split()))
s = input()
a = []
for i in range(len(s)):
if s[i] == '*': a.append(i)
if len(a) < 3: print(len(a))
else:
i = 0
count = 0
while i < len(a) - 1:
j = i
while j < len(a) and a[j] <= a[i] + k: j += 1
count += 1
i = j-1
print(count+1)
``` | instruction | 0 | 71,892 | 0 | 143,784 |
Yes | output | 1 | 71,892 | 0 | 143,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
i = s.find('*')
ans = 1
while True:
j = min(n - 1, i + k)
while i < j and s[j] == '.':
j -= 1
if i == j:
break
ans += 1
i = j
print(ans)
``` | instruction | 0 | 71,893 | 0 | 143,786 |
Yes | output | 1 | 71,893 | 0 | 143,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
import math
import sys
from collections import *
import itertools
def cint() : return list(map(int, sys.stdin.readline().strip().split()))
def cstr() : return list(map(str, input().split(' ')))
def gcd(a,b):
if (b == 0): return a
return gcd(b, a%b)
def solve(t):
n,k = cint()
lst = [i for i in input()]
counter = 0
sindx = -1
eindx = -1
for i in range(n):
if lst[i]== '*':
lst[i] = 'x'
counter += 1
sindx = i
break
for i in range(n-1,-1,-1):
if lst[i] == '*':
lst[i] = 'x'
counter += 1
eindx = i
break
indx = sindx
stars = []
for i in range(sindx+1,n):
if lst[i] == '*':
if i-indx < k: stars.append(i)
elif i-indx==k:
lst[i] = 'x'
indx = i
counter += 1
else:
for j in range(len(stars)-1,-1,-1):
tindx = stars[j]
if tindx!=-1:
if tindx-indx <= k:
lst[tindx] = 'x'
stars[j] = -1
indx = tindx
counter += 1
break
stars.append(i)
# print(lst)
for i in range(eindx-1,-1,-1):
if lst[i]=='x':
# print(i, 'aaa')
if abs(eindx - i) >k:
counter += 1
break
print(counter)
if __name__ == "__main__":
t = int(input())
# t =1
for i in range(1,t+1):
solve(i)
``` | instruction | 0 | 71,894 | 0 | 143,788 |
Yes | output | 1 | 71,894 | 0 | 143,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
#author: anshul_129
from sys import stdin, stdout, maxsize
from math import sqrt, log, factorial as ft, gcd, ceil
from collections import defaultdict as D
from bisect import insort
for _ in range(int(input())):
n, k = map(int, stdin.readline().strip().split())
s = stdin.readline().strip()
#a = list(map(int, stdin.readline().strip().split()))
#b = list(map(int, stdin.readline().strip().split()))
#d = D(lambda: 0)
#print("Case #" + str(_ + 1) + ": ", end = '')
c = s.count('*')
if c <= 2: print(c)
else:
fi = s.index('*')
li = n - 1 - s[::-1].index('*')
ans = 2
dist = 0
for i in range(fi + 1, li + 1):
dist += 1
if dist > k:
for j in range(i, fi, -1):
if s[j] == '*':
prev = j
ans += 1
break
dist = i - j
print(ans)
``` | instruction | 0 | 71,895 | 0 | 143,790 |
No | output | 1 | 71,895 | 0 | 143,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
def GetList(): return list(map(int, input().split()))
from collections import defaultdict
def ceil(n):
if int(n)-n ==0:
n = int(n)
else:
n = int(n)+1
return n
def main():
t = int(input())
for num in range(t):
n, k = map(int, input().split())
s = input()
for i in range(n):
if s[i]=="*":
break
for k in range(n):
if s[n-k-1]:
if s[k]=='*':
break
if (n-k-1)==i:
print(1)
elif (n-k-1)-i<2:
print(1)
else:
m = ((n-k-1)-i)//3 + 2
print(m)
main()
``` | instruction | 0 | 71,896 | 0 | 143,792 |
No | output | 1 | 71,896 | 0 | 143,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
s = list(input())
a = 0
c = 0
for i in range(n):
if s[i] == "*":
a = i
break
s[a] = "x"
c += 1
a += k
while a < n:
if s[a] == "*":
s[a] = "x"
c += 1
a += k
if s[len(s) - 1] == "*":
# s[len(s) - a] = "x"
c += 1
print(c)
# print(s)
``` | instruction | 0 | 71,897 | 0 | 143,794 |
No | output | 1 | 71,897 | 0 | 143,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a number k and a string s of length n, consisting of the characters '.' and '*'. You want to replace some of the '*' characters with 'x' characters so that the following conditions are met:
* The first character '*' in the original string should be replaced with 'x';
* The last character '*' in the original string should be replaced with 'x';
* The distance between two neighboring replaced characters 'x' must not exceed k (more formally, if you replaced characters at positions i and j (i < j) and at positions [i+1, j-1] there is no "x" symbol, then j-i must be no more than k).
For example, if n=7, s=.**.*** and k=3, then the following strings will satisfy the conditions above:
* .xx.*xx;
* .x*.x*x;
* .xx.xxx.
But, for example, the following strings will not meet the conditions:
* .**.*xx (the first character '*' should be replaced with 'x');
* .x*.xx* (the last character '*' should be replaced with 'x');
* .x*.*xx (the distance between characters at positions 2 and 6 is greater than k=3).
Given n, k, and s, find the minimum number of '*' characters that must be replaced with 'x' in order to meet the above conditions.
Input
The first line contains one integer t (1 ≤ t ≤ 500). Then t test cases follow.
The first line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 50).
The second line of each test case contains a string s of length n, consisting of the characters '.' and '*'.
It is guaranteed that there is at least one '*' in the string s.
It is guaranteed that the distance between any two neighboring '*' characters does not exceed k.
Output
For each test case output the minimum number of '*' characters that must be replaced with 'x' characters in order to satisfy the conditions above.
Example
Input
5
7 3
.**.***
5 1
..*..
5 2
*.*.*
3 2
*.*
1 1
*
Output
3
1
3
2
1
Submitted Solution:
```
for test in range(int(input())):
n, k = map(int, input().split())
s = input().strip()
if s.count('*') >= 2:
s = s[::-1]
s = s.replace('*', 'x', 1)
s = s[::-1]
s = s.replace('*', 'x', 1)
i = s.index('x')
j = n - s[::-1].index('x')
while i < j:
if 'x' not in s[i + 1:i + k + 1]:
temp = s[i + 1:i + k + 1][::-1]
s = s[:i + 1] + temp.replace('*', 'x', 1) + s[i + k + 1:]
i += s.index('x') + 1
else:
i += k
print(s.count('x'))
``` | instruction | 0 | 71,898 | 0 | 143,796 |
No | output | 1 | 71,898 | 0 | 143,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an English class Nick had nothing to do at all, and remembered about wonderful strings called palindromes. We should remind you that a string is called a palindrome if it can be read the same way both from left to right and from right to left. Here are examples of such strings: «eye», «pop», «level», «aba», «deed», «racecar», «rotor», «madam».
Nick started to look carefully for all palindromes in the text that they were reading in the class. For each occurrence of each palindrome in the text he wrote a pair — the position of the beginning and the position of the ending of this occurrence in the text. Nick called each occurrence of each palindrome he found in the text subpalindrome. When he found all the subpalindromes, he decided to find out how many different pairs among these subpalindromes cross. Two subpalindromes cross if they cover common positions in the text. No palindrome can cross itself.
Let's look at the actions, performed by Nick, by the example of text «babb». At first he wrote out all subpalindromes:
• «b» — 1..1 • «bab» — 1..3 • «a» — 2..2 • «b» — 3..3 • «bb» — 3..4 • «b» — 4..4
Then Nick counted the amount of different pairs among these subpalindromes that cross. These pairs were six:
1. 1..1 cross with 1..3 2. 1..3 cross with 2..2 3. 1..3 cross with 3..3 4. 1..3 cross with 3..4 5. 3..3 cross with 3..4 6. 3..4 cross with 4..4
Since it's very exhausting to perform all the described actions manually, Nick asked you to help him and write a program that can find out the amount of different subpalindrome pairs that cross. Two subpalindrome pairs are regarded as different if one of the pairs contains a subpalindrome that the other does not.
Input
The first input line contains integer n (1 ≤ n ≤ 2·106) — length of the text. The following line contains n lower-case Latin letters (from a to z).
Output
In the only line output the amount of different pairs of two subpalindromes that cross each other. Output the answer modulo 51123987.
Examples
Input
4
babb
Output
6
Input
2
aa
Output
2
Submitted Solution:
```
n=int(input())
x=input()
s=0
for i in range(0,n):
for j in range(i+1,n+1):
t=x[i:j]
if t==t[::-1]:
s+=1
if x==x[::-1]:
s-=1
print(s)
``` | instruction | 0 | 71,931 | 0 | 143,862 |
No | output | 1 | 71,931 | 0 | 143,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an English class Nick had nothing to do at all, and remembered about wonderful strings called palindromes. We should remind you that a string is called a palindrome if it can be read the same way both from left to right and from right to left. Here are examples of such strings: «eye», «pop», «level», «aba», «deed», «racecar», «rotor», «madam».
Nick started to look carefully for all palindromes in the text that they were reading in the class. For each occurrence of each palindrome in the text he wrote a pair — the position of the beginning and the position of the ending of this occurrence in the text. Nick called each occurrence of each palindrome he found in the text subpalindrome. When he found all the subpalindromes, he decided to find out how many different pairs among these subpalindromes cross. Two subpalindromes cross if they cover common positions in the text. No palindrome can cross itself.
Let's look at the actions, performed by Nick, by the example of text «babb». At first he wrote out all subpalindromes:
• «b» — 1..1 • «bab» — 1..3 • «a» — 2..2 • «b» — 3..3 • «bb» — 3..4 • «b» — 4..4
Then Nick counted the amount of different pairs among these subpalindromes that cross. These pairs were six:
1. 1..1 cross with 1..3 2. 1..3 cross with 2..2 3. 1..3 cross with 3..3 4. 1..3 cross with 3..4 5. 3..3 cross with 3..4 6. 3..4 cross with 4..4
Since it's very exhausting to perform all the described actions manually, Nick asked you to help him and write a program that can find out the amount of different subpalindrome pairs that cross. Two subpalindrome pairs are regarded as different if one of the pairs contains a subpalindrome that the other does not.
Input
The first input line contains integer n (1 ≤ n ≤ 2·106) — length of the text. The following line contains n lower-case Latin letters (from a to z).
Output
In the only line output the amount of different pairs of two subpalindromes that cross each other. Output the answer modulo 51123987.
Examples
Input
4
babb
Output
6
Input
2
aa
Output
2
Submitted Solution:
```
num = int(input())
p = input()
v = 0
for i in range(0, num):
for j in range(i+1, num+1):
a = p[i:j]
if a == a[::-1]:
v += 1
print(v)
``` | instruction | 0 | 71,932 | 0 | 143,864 |
No | output | 1 | 71,932 | 0 | 143,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an English class Nick had nothing to do at all, and remembered about wonderful strings called palindromes. We should remind you that a string is called a palindrome if it can be read the same way both from left to right and from right to left. Here are examples of such strings: «eye», «pop», «level», «aba», «deed», «racecar», «rotor», «madam».
Nick started to look carefully for all palindromes in the text that they were reading in the class. For each occurrence of each palindrome in the text he wrote a pair — the position of the beginning and the position of the ending of this occurrence in the text. Nick called each occurrence of each palindrome he found in the text subpalindrome. When he found all the subpalindromes, he decided to find out how many different pairs among these subpalindromes cross. Two subpalindromes cross if they cover common positions in the text. No palindrome can cross itself.
Let's look at the actions, performed by Nick, by the example of text «babb». At first he wrote out all subpalindromes:
• «b» — 1..1 • «bab» — 1..3 • «a» — 2..2 • «b» — 3..3 • «bb» — 3..4 • «b» — 4..4
Then Nick counted the amount of different pairs among these subpalindromes that cross. These pairs were six:
1. 1..1 cross with 1..3 2. 1..3 cross with 2..2 3. 1..3 cross with 3..3 4. 1..3 cross with 3..4 5. 3..3 cross with 3..4 6. 3..4 cross with 4..4
Since it's very exhausting to perform all the described actions manually, Nick asked you to help him and write a program that can find out the amount of different subpalindrome pairs that cross. Two subpalindrome pairs are regarded as different if one of the pairs contains a subpalindrome that the other does not.
Input
The first input line contains integer n (1 ≤ n ≤ 2·106) — length of the text. The following line contains n lower-case Latin letters (from a to z).
Output
In the only line output the amount of different pairs of two subpalindromes that cross each other. Output the answer modulo 51123987.
Examples
Input
4
babb
Output
6
Input
2
aa
Output
2
Submitted Solution:
```
from itertools import combinations
def palindrome(a):
s=a[::-1]
if(s==a):
return 1
else:
return 0
n=int(input())
s=input()
c=0
j=2
if(s==s[::-1]):
c=c+1
for i in range(n):
l=list(combinations(s,j))
for a in l:
p = ''.join(a)
if (palindrome(p) == 1):
c = c + 1
j=j+1
print(c)
``` | instruction | 0 | 71,933 | 0 | 143,866 |
No | output | 1 | 71,933 | 0 | 143,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In an English class Nick had nothing to do at all, and remembered about wonderful strings called palindromes. We should remind you that a string is called a palindrome if it can be read the same way both from left to right and from right to left. Here are examples of such strings: «eye», «pop», «level», «aba», «deed», «racecar», «rotor», «madam».
Nick started to look carefully for all palindromes in the text that they were reading in the class. For each occurrence of each palindrome in the text he wrote a pair — the position of the beginning and the position of the ending of this occurrence in the text. Nick called each occurrence of each palindrome he found in the text subpalindrome. When he found all the subpalindromes, he decided to find out how many different pairs among these subpalindromes cross. Two subpalindromes cross if they cover common positions in the text. No palindrome can cross itself.
Let's look at the actions, performed by Nick, by the example of text «babb». At first he wrote out all subpalindromes:
• «b» — 1..1 • «bab» — 1..3 • «a» — 2..2 • «b» — 3..3 • «bb» — 3..4 • «b» — 4..4
Then Nick counted the amount of different pairs among these subpalindromes that cross. These pairs were six:
1. 1..1 cross with 1..3 2. 1..3 cross with 2..2 3. 1..3 cross with 3..3 4. 1..3 cross with 3..4 5. 3..3 cross with 3..4 6. 3..4 cross with 4..4
Since it's very exhausting to perform all the described actions manually, Nick asked you to help him and write a program that can find out the amount of different subpalindrome pairs that cross. Two subpalindrome pairs are regarded as different if one of the pairs contains a subpalindrome that the other does not.
Input
The first input line contains integer n (1 ≤ n ≤ 2·106) — length of the text. The following line contains n lower-case Latin letters (from a to z).
Output
In the only line output the amount of different pairs of two subpalindromes that cross each other. Output the answer modulo 51123987.
Examples
Input
4
babb
Output
6
Input
2
aa
Output
2
Submitted Solution:
```
n=int(input())
x=input()
s=0
for i in range(0,n):
for j in range(i+1,n+1):
t=x[i:j]
if t==t[::-1]:
s+=1
print(s)
``` | instruction | 0 | 71,934 | 0 | 143,868 |
No | output | 1 | 71,934 | 0 | 143,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0). | instruction | 0 | 72,056 | 0 | 144,112 |
Tags: greedy, hashing, string suffix structures, strings
Correct Solution:
```
from sys import stdin, stdout, setrecursionlimit
input = stdin.readline
import string
characters = string.ascii_uppercase
from random import choice
# digits = string.digits
# setrecursionlimit(int(1e6))
# dir = [-1,0,1,0,-1]
# moves = 'NESW'
inf = float('inf')
from functools import cmp_to_key
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')
mod = int(1e9+9)
def binary(a,b):
res = 1
while b:
if b&1:
res *= a
res %= mod
a*=a
a%=mod
b//=2
return res
def solve():
n, m = geti()
ans = [choice(characters) for _ in range(n)]
s = gets()
k = len(s)
a = getl()
a = {i-1: True for i in a}
mod = int(1e9+9)
index = k
for i in range(n):
if i in a:
index = 0
if index < k:
ans[i] = s[index]
index += 1
p = 31
pow = 1
hash = 0
res = 0
# print(ans, res)
for i in range(k):
hash += (ord(s[i])-ord('a') + 1)*pow
res += (ord(ans[i])-ord('a') + 1)*pow
hash %= mod
res %= mod
if i!=k-1:
pow *= p
pow %= mod
if 0 in a:
if hash != res:
return 0
div = binary(p, mod-2)
for i in range(k,n):
res -= (ord(ans[i-k]) - ord('a') + 1)
res *= div
res %= mod
res += (ord(ans[i])-ord('a') + 1)*pow
res %= mod
if i-k+1 in a and res != hash:
return 0
res = 1
mod = int(1e9+7)
# print(ans)
for i in ans:
if i.isupper():
res *= 26
res %= mod
return res
if __name__=='__main__':
print(solve())
``` | output | 1 | 72,056 | 0 | 144,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0). | instruction | 0 | 72,057 | 0 | 144,114 |
Tags: greedy, hashing, string suffix structures, strings
Correct Solution:
```
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")
##########################################################
from collections import Counter,defaultdict
import math
#for _ in range(int(input())):
#n=int(input())
def matching(s):
le = len(s)
pi=[0]*le
for i in range(1,le):
j=pi[i-1]
while j>0 and s[i]!=s[j]:
j=pi[j-1]
if(s[i]==s[j]):
j+=1
pi[i]=j
#return pi
## to return list of values
w=set()
i=le-1
while pi[i]!=0:
w.add(le-pi[i])
i=pi[i]-1
return w
n,m=map(int,input().split())
p=input()
if m==0:
print(pow(26,n,10**9+7))
sys.exit()
arr=list(map(int, input().split()))
#a1=list(map(int, input().split()))
ans=0
l=len(p)
pre=matching(p)
filled=l
for i in range(1,m):
if (arr[i]-arr[i-1]<l and (arr[i]-arr[i-1]) not in pre):
print(0)
sys.exit()
filled+=min(l,arr[i]-arr[i-1])
print(pow(26,n-filled,10**9+7))
``` | output | 1 | 72,057 | 0 | 144,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0). | instruction | 0 | 72,058 | 0 | 144,116 |
Tags: greedy, hashing, string suffix structures, strings
Correct Solution:
```
# -*- coding: utf-8 -*-
def solve():
n, m = map(int, input().split())
p = input()
if m == 0:
return powmod(n)
delta = len(p) - 1
ys = map(int, input().split())
tail = 0
free_chars = 0
for y in ys:
if y > tail:
free_chars += y - tail - 1
elif not is_consistent(p, tail - y + 1):
return 0
tail = y + delta
free_chars += n - tail
return powmod(free_chars)
ok_set = set()
def is_consistent(p, margin):
global ok_set
if margin in ok_set:
return True
elif p[:margin] == p[-margin:]:
ok_set.add(margin)
return True
else:
return False
def powmod(p):
mod = 10**9 + 7
pbin = bin(p)[2:][-1::-1]
result = 26 if pbin[0] == '1' else 1
tmp = 26
for bit in pbin[1:]:
tmp *= tmp
tmp %= mod
if bit == '1':
result *= tmp
result %= mod
return result
print(solve())
``` | output | 1 | 72,058 | 0 | 144,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0). | instruction | 0 | 72,059 | 0 | 144,118 |
Tags: greedy, hashing, string suffix structures, strings
Correct Solution:
```
import sys
def prefix(s):
m = len(s)
v = [0]*len(s)
for i in range(1,len(s)):
k = v[i-1]
while k > 0 and s[k] != s[i]:
k = v[k-1]
if s[k] == s[i]:
k = k + 1
v[i] = k
w = set()
i = m-1
while v[i] != 0:
w.add(m-v[i])
i = v[i]-1
return w
n,m = map(int, input().split())
if m == 0:
print(pow(26, n, 1000000007))
sys.exit(0)
p = input()
l = len(p)
x = list(map(int,input().split()))
w = prefix(p)
busy = l
for i in range(1,m):
if x[i]-x[i-1] < l and (x[i] - x[i-1]) not in w:
print(0)
sys.exit(0)
busy += min(x[i]-x[i-1], l)
print(pow(26,n-busy, 1000000007))
``` | output | 1 | 72,059 | 0 | 144,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0). | instruction | 0 | 72,060 | 0 | 144,120 |
Tags: greedy, hashing, string suffix structures, strings
Correct Solution:
```
def preZ(s): #preprocessing by Z algo
n = len(s)
z = [0]*n
z[0] = n
r = 0
if n==1: return z
while r+1<n and s[r]==s[r+1]: r+=1
z[1] = r #note z=length! not 0-indexed
l = 1 if r>0 else 0
for k in range(2,n):
bl = r+1-k #|\beta|
gl = z[k-l] #|\gamma|
if gl<bl:
z[k]=z[k-l] #Case2a
else:
j=max(0,r-k+1) #Case1 & 2b
while k+j<n and s[j]==s[k+j]: j+=1
z[k]=j
l,r =k,k+j-1
return z
pp = int(1e9)+7
def binpow(b,e):
r = 1
while True:
if e &1: r=(r*b)%pp
e = e>>1
if e==0: break
b = (b*b)%pp
return r
def f(p,l,n): #pattern, match list, size of text
m = len(p)
if len(l)==0:
return binpow(26,n)
z = preZ(p)
s = set([i for i in range(m) if z[i]+i==m])
fc = l[0]-1
for i in range(1,len(l)):
r = l[i-1]+m
if l[i]>r:
fc += l[i]-r
continue
if l[i]<r and l[i]-l[i-1] not in s:
return 0
fc += n-(l[-1]+m-1)
return binpow(26,fc)
n,m = list(map(int,input().split()))
p = input()
l = list(map(int,input().split())) if m>0 else []
print(f(p,l,n))
``` | output | 1 | 72,060 | 0 | 144,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0). | instruction | 0 | 72,061 | 0 | 144,122 |
Tags: greedy, hashing, string suffix structures, strings
Correct Solution:
```
from sys import stdin, stdout, setrecursionlimit
input = stdin.readline
# digits = string.digits
# setrecursionlimit(int(1e6))
# dir = [-1,0,1,0,-1]
# moves = 'NESW'
inf = float('inf')
from functools import cmp_to_key
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')
mod = int(1e9+9)
def binary(a,b):
res = 1
while b:
if b&1:
res *= a
res %= mod
a*=a
a%=mod
b//=2
return res
def solve():
n, m = geti()
ans = ['A']*n
s = gets()
k = len(s)
a = getl()
a = {i-1: True for i in a}
mod = int(1e9+9)
index = k
for i in range(n):
if i in a:
index = 0
if index < k:
ans[i] = s[index]
index += 1
p = 31
pow = 1
hash = 0
res = 0
# print(ans, res)
for i in range(k):
hash += (ord(s[i])-ord('a') + 1)*pow
res += (ord(ans[i])-ord('a') + 1)*pow
hash %= mod
res %= mod
if i!=k-1:
pow *= p
pow %= mod
if 0 in a:
if hash != res:
return 0
div = binary(p, mod-2)
for i in range(k,n):
res -= (ord(ans[i-k]) - ord('a') + 1)
res *= div
res %= mod
res += (ord(ans[i])-ord('a') + 1)*pow
res %= mod
if i-k+1 in a and res != hash:
return 0
res = 1
mod = int(1e9+7)
# print(ans)
for i in ans:
if i == 'A':
res *= 26
res %= mod
return res
if __name__=='__main__':
print(solve())
``` | output | 1 | 72,061 | 0 | 144,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0).
Submitted Solution:
```
import sys
def prefix(s):
m = len(s)
v = [0]*len(s)
for i in range(1,len(s)):
k = v[i-1]
while k > 0 and s[k] != s[i]:
k = v[k-1]
if s[k] == s[i]:
k = k + 1
v[i] = k
w = set()
i = m-1
while v[i] != 0:
w.add(m-v[i])
i = v[i]-1
return w
n,m = map(int, input().split())
p = input()
l = len(p)
x = list(map(int,input().split()))
w = prefix(p)
busy = l
for i in range(1,m):
if (x[i] - x[i-1]) not in w:
print(0)
sys.exit(0)
busy += min(x[i]-x[i-1], l)
print(pow(26,n-busy, 1000000007))
``` | instruction | 0 | 72,062 | 0 | 144,124 |
No | output | 1 | 72,062 | 0 | 144,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0).
Submitted Solution:
```
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")
##########################################################
from collections import Counter,defaultdict
import math
#for _ in range(int(input())):
#n=int(input())
def matching(s):
le = len(s)
pi=[0]*le
for i in range(1,le):
j=pi[i-1]
while j>0 and s[i]!=s[j]:
j=pi[j-1]
if(s[i]==s[j]):
j+=1
pi[i]=j
#return pi
## to return list of values
w=set()
i=le-1
while pi[i]!=0:
w.add(le-pi[i])
i=pi[i]-1
return w
n,m=map(int,input().split())
p=input()
if m==0:
print(pow(26,n,10**9+7))
sys.exit()
arr=list(map(int, input().split()))
#a1=list(map(int, input().split()))
ans=0
l=len(p)
pre=matching(p)
filled=l
for i in range(1,m):
if (arr[i]-arr[i-1]<l and (arr[i]-arr[i-1]) not in pre):
print(pre)
print(0)
sys.exit()
filled+=min(l,arr[i]-arr[i-1])
print(pow(26,n-filled,10**9+7))
``` | instruction | 0 | 72,063 | 0 | 144,126 |
No | output | 1 | 72,063 | 0 | 144,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0).
Submitted Solution:
```
import sys
def prefix(s):
m = len(s)
v = [0]*len(s)
for i in range(1,len(s)):
k = v[i-1]
while k > 0 and s[k] != s[i]:
k = v[k-1]
if s[k] == s[i]:
k = k + 1
v[i] = k
w = set()
i = m-1
while v[i] != 0:
w.add(m-v[i])
i = v[i]-1
return w
n,m = map(int, input().split())
p = input()
l = len(p)
if m != 0:
x = list(map(int,input().split()))
w = prefix(p)
busy = l
for i in range(1,m):
if x[i]-x[i-1] < l and (x[i] - x[i-1]) not in w:
print(0)
sys.exit(0)
busy += min(x[i]-x[i-1], l)
print(pow(26,n-busy, 1000000007))
``` | instruction | 0 | 72,064 | 0 | 144,128 |
No | output | 1 | 72,064 | 0 | 144,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.
<image>
Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x1 < x2 < ... < xk where p matches s. More formally, for each xi (1 ≤ i ≤ k) he condition sxisxi + 1... sxi + |p| - 1 = p is fullfilled.
Then Malekas wrote down one of subsequences of x1, x2, ... xk (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).
After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.
Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.
Answer can be very large, so Tavas wants you to print the answer modulo 109 + 7.
Input
The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 106 and 0 ≤ m ≤ n - |p| + 1).
The second line contains string p (1 ≤ |p| ≤ n).
The next line contains m space separated integers y1, y2, ..., ym, Malekas' subsequence (1 ≤ y1 < y2 < ... < ym ≤ n - |p| + 1).
Output
In a single line print the answer modulo 1000 000 007.
Examples
Input
6 2
ioi
1 3
Output
26
Input
5 2
ioi
1 2
Output
0
Note
In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.
Here |x| denotes the length of string x.
Please note that it's possible that there is no such string (answer is 0).
Submitted Solution:
```
from sys import stdin, stdout, setrecursionlimit
input = stdin.readline
# import string
# characters = string.ascii_lowercase
# digits = string.digits
# setrecursionlimit(int(1e6))
# dir = [-1,0,1,0,-1]
# moves = 'NESW'
inf = float('inf')
from functools import cmp_to_key
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')
mod = int(1e9+9)
def binary(a,b):
res = 1
while b:
if b&1:
res *= a
res %= mod
a*=a
a%=mod
b//=2
return res
def solve():
n, m = geti()
ans = ['A']*n
s = gets()
k = len(s)
a = Counter(getl())
mod = int(1e9+9)
index = k
for i in range(n):
if i in a:
index = 0
if index < k:
ans[i] = s[index]
index += 1
p = 31
pow = 1
hash = 0
res = 0
# print(ans, res)
for i in range(k):
hash += (ord(s[i])-ord('a') + 1)*pow
res += (ord(ans[i])-ord('a') + 1)*pow
hash %= mod
res %= mod
if i!=k-1:
pow *= p
pow %= mod
if 0 in a:
if hash != res:
return 0
div = binary(p, mod-2)
for i in range(k,n):
res -= (ord(ans[i-k]) - ord('a') + 1)
res *= div
res %= mod
res += (ord(ans[i])-ord('a') + 1)*pow
res %= mod
if i-k+1 in a and res != hash:
return 0
res = 1
mod = int(1e9+7)
for i in ans:
if i == 'A':
res *= 26
res %= mod
return res
if __name__=='__main__':
print(solve())
``` | instruction | 0 | 72,065 | 0 | 144,130 |
No | output | 1 | 72,065 | 0 | 144,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A string a of length m is called antipalindromic iff m is even, and for each i (1 ≤ i ≤ m) ai ≠ am - i + 1.
Ivan has a string s consisting of n lowercase Latin letters; n is even. He wants to form some string t that will be an antipalindromic permutation of s. Also Ivan has denoted the beauty of index i as bi, and the beauty of t as the sum of bi among all indices i such that si = ti.
Help Ivan to determine maximum possible beauty of t he can get.
Input
The first line contains one integer n (2 ≤ n ≤ 100, n is even) — the number of characters in s.
The second line contains the string s itself. It consists of only lowercase Latin letters, and it is guaranteed that its letters can be reordered to form an antipalindromic string.
The third line contains n integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the beauty of index i.
Output
Print one number — the maximum possible beauty of t.
Examples
Input
8
abacabac
1 1 1 1 1 1 1 1
Output
8
Input
8
abaccaba
1 2 3 4 5 6 7 8
Output
26
Input
8
abacabca
1 2 3 4 4 3 2 1
Output
17 | instruction | 0 | 72,230 | 0 | 144,460 |
Tags: flows, graphs, greedy
Correct Solution:
```
#https://pymotw.com/2/collections/counter.html
#same code as mmaxio
from collections import Counter
r = lambda: map(int, input().split())
def main():
n, = r()
s = input()
cost = list(r())
ans = 0
cnt = Counter()
for i in range(n // 2):
if s[i] == s[n - 1 - i]:
ans += min(cost[i], cost[n - 1 - i])
cnt[s[i]] += 1
total = sum(cnt.values())
if total > 0:
ch, occ = cnt.most_common(1)[0]
avail = []
if occ > total - occ:# if highest occurence is more than the 50% of total then we will look for the letters which does not have pairs and are not equal to the letter with the highest ocuurence
for i in range(n // 2):
if s[i] != s[n - 1 - i] and s[i] != ch and s[n - 1 - i] != ch:
avail.append(min(cost[i], cost[n - 1 - i]))
avail.sort()
ans += sum(avail[:2 * occ - total])
print(sum(cost)-ans)
main()
#suppose total is 100 and highest occ is 51...difference between highest occ and remaining can be found using this form 2*occ-total as it is a simplified form of two steps 1.total-occ=remaining and 2.occ-remaining which is this case is 2 if highest occ is <= 50 % of total then it can be satisfied by remaining 50% but if it is greater than 50% then we have to use the letters of of the total
``` | output | 1 | 72,230 | 0 | 144,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string a of length m is called antipalindromic iff m is even, and for each i (1 ≤ i ≤ m) ai ≠ am - i + 1.
Ivan has a string s consisting of n lowercase Latin letters; n is even. He wants to form some string t that will be an antipalindromic permutation of s. Also Ivan has denoted the beauty of index i as bi, and the beauty of t as the sum of bi among all indices i such that si = ti.
Help Ivan to determine maximum possible beauty of t he can get.
Input
The first line contains one integer n (2 ≤ n ≤ 100, n is even) — the number of characters in s.
The second line contains the string s itself. It consists of only lowercase Latin letters, and it is guaranteed that its letters can be reordered to form an antipalindromic string.
The third line contains n integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the beauty of index i.
Output
Print one number — the maximum possible beauty of t.
Examples
Input
8
abacabac
1 1 1 1 1 1 1 1
Output
8
Input
8
abaccaba
1 2 3 4 5 6 7 8
Output
26
Input
8
abacabca
1 2 3 4 4 3 2 1
Output
17
Submitted Solution:
```
k=int(input())
s=list(input())
beauty=list(map(int, input().split(" ")))
for i in range(k):
if s[i]==s[k-i-1]:
if beauty[i]<=beauty[k-i-1]:
mini=111
pos=i
for b in range(len(beauty)//2):
if s[b]!=s[i] and mini>beauty[b]:
mini=beauty[b]
pos=b
temp=s[pos]
s[pos]=s[i]
s[i]=temp
beauty[i]=0
beauty[pos]=0
else:
for b in range(len(beauty)//2,len(beauty)):
mini=111
pos=i
if beauty[b]<beauty[i] and s[b]!=s[i] and mini>beauty[b]:
mini=beauty[b]
pos=b
temp=s[pos]
s[pos]=s[i]
s[i]=temp
beauty[i]=0
beauty[pos]=0
print(sum(beauty))
``` | instruction | 0 | 72,233 | 0 | 144,466 |
No | output | 1 | 72,233 | 0 | 144,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string a of length m is called antipalindromic iff m is even, and for each i (1 ≤ i ≤ m) ai ≠ am - i + 1.
Ivan has a string s consisting of n lowercase Latin letters; n is even. He wants to form some string t that will be an antipalindromic permutation of s. Also Ivan has denoted the beauty of index i as bi, and the beauty of t as the sum of bi among all indices i such that si = ti.
Help Ivan to determine maximum possible beauty of t he can get.
Input
The first line contains one integer n (2 ≤ n ≤ 100, n is even) — the number of characters in s.
The second line contains the string s itself. It consists of only lowercase Latin letters, and it is guaranteed that its letters can be reordered to form an antipalindromic string.
The third line contains n integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the beauty of index i.
Output
Print one number — the maximum possible beauty of t.
Examples
Input
8
abacabac
1 1 1 1 1 1 1 1
Output
8
Input
8
abaccaba
1 2 3 4 5 6 7 8
Output
26
Input
8
abacabca
1 2 3 4 4 3 2 1
Output
17
Submitted Solution:
```
#https://pymotw.com/2/collections/counter.html
#same code as mmaxio
from collections import Counter
r = lambda: map(int, input().split())
def main():
n, = r()
s = input()
cost = list(r())
ans = 0
cnt = Counter()
for i in range(n // 2):
if s[i] == s[n - 1 - i]:
ans += min(cost[i], cost[n - 1 - i])
cnt[s[i]] += 1
total = sum(cnt.values())
if total > 0:
ch, occ = cnt.most_common(1)[0]
avail = []
if occ > total - occ:# if highest occurence is more than the 50% of total then we will look for the letters which does not have pairs and are not equal to the letter with the highest ocuurence
for i in range(n // 2):
if s[i] != s[n - 1 - i] and s[i] != ch and s[n - 1 - i] != ch:
avail.append(min(cost[i], cost[n - 1 - i]))
avail=sorted(avail,reverse=True)
ans += sum(avail[:2 * occ - total])
print(sum(cost)-ans)
main()
#suppose total is 100 and highest occ is 51...difference between highest occ and remaining can be found using this form 2*occ-total as it is a simplified form of two steps 1.total-occ=remaining and 2.occ-remaining which is this case is 2 if highest occ is <= 50 % of total then it can be satisfied by remaining 50% but if it is greater than 50% then we have to use the letters of of the total
``` | instruction | 0 | 72,234 | 0 | 144,468 |
No | output | 1 | 72,234 | 0 | 144,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string a of length m is called antipalindromic iff m is even, and for each i (1 ≤ i ≤ m) ai ≠ am - i + 1.
Ivan has a string s consisting of n lowercase Latin letters; n is even. He wants to form some string t that will be an antipalindromic permutation of s. Also Ivan has denoted the beauty of index i as bi, and the beauty of t as the sum of bi among all indices i such that si = ti.
Help Ivan to determine maximum possible beauty of t he can get.
Input
The first line contains one integer n (2 ≤ n ≤ 100, n is even) — the number of characters in s.
The second line contains the string s itself. It consists of only lowercase Latin letters, and it is guaranteed that its letters can be reordered to form an antipalindromic string.
The third line contains n integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the beauty of index i.
Output
Print one number — the maximum possible beauty of t.
Examples
Input
8
abacabac
1 1 1 1 1 1 1 1
Output
8
Input
8
abaccaba
1 2 3 4 5 6 7 8
Output
26
Input
8
abacabca
1 2 3 4 4 3 2 1
Output
17
Submitted Solution:
```
from collections import Counter
r = lambda: map(int, input().split())
def main():
n, = r()
s = input()
cost = list(r())
ans = 0
cnt = Counter()
for i in range(n // 2):
if s[i] == s[n - 1 - i]:
ans += min(cost[i], cost[n - 1 - i])
cnt[s[i]] += 1
total = sum(cnt.values())
if total > 0:
ch, occ = cnt.most_common(1)[0]
avail = []
if occ > total - occ:
for i in range(n // 2):
if s[i] != s[n - 1 - i] and s[i] != ch and s[n - 1 - i] != ch:
avail.append(min(cost[i], cost[n - 1 - i]))
avail.sort(reverse=True)
ans += sum(avail[:2 * occ - total])
print(sum(cost) - ans)
main()
``` | instruction | 0 | 72,235 | 0 | 144,470 |
No | output | 1 | 72,235 | 0 | 144,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A string a of length m is called antipalindromic iff m is even, and for each i (1 ≤ i ≤ m) ai ≠ am - i + 1.
Ivan has a string s consisting of n lowercase Latin letters; n is even. He wants to form some string t that will be an antipalindromic permutation of s. Also Ivan has denoted the beauty of index i as bi, and the beauty of t as the sum of bi among all indices i such that si = ti.
Help Ivan to determine maximum possible beauty of t he can get.
Input
The first line contains one integer n (2 ≤ n ≤ 100, n is even) — the number of characters in s.
The second line contains the string s itself. It consists of only lowercase Latin letters, and it is guaranteed that its letters can be reordered to form an antipalindromic string.
The third line contains n integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the beauty of index i.
Output
Print one number — the maximum possible beauty of t.
Examples
Input
8
abacabac
1 1 1 1 1 1 1 1
Output
8
Input
8
abaccaba
1 2 3 4 5 6 7 8
Output
26
Input
8
abacabca
1 2 3 4 4 3 2 1
Output
17
Submitted Solution:
```
class letter(object):
def __init__(self,let,val):
self.let=let
self.val=val
def __lt__(self,other):
return self.val<other.val
n=int(input())
s=input()
candi=[[] for i in range(n//2)]
ans=0
for i,vl in enumerate(map(int,input().split())):
candi[min(i,n-i-1)].append((letter)(s[i],vl))
ans+=vl
for i in range(n//2):
candi[i].sort()
ti=[0 for i in range(26)]
sum=0
for i in range(n//2):
if candi[i][0].let==candi[i][1].let:
ans-=candi[i][0].val
ti[ord(candi[i][0].let)-ord('a')]+=1
sum+=1
mx=0
p=0
for i in range(26):
if ti[i]>mx:
mx=ti[i]
p=i
b=[]
for i in range(n//2):
if ord(candi[i][0].let)-ord('a')!=p:
b.append(candi[i][0])
elif ord(candi[i][1].let)-ord('a')!=p:
b.append(candi[i][1])
b.sort()
i=0
while mx*2>sum:
sum+=1
ans-=b[i].val
i+=1
print(ans)
``` | instruction | 0 | 72,236 | 0 | 144,472 |
No | output | 1 | 72,236 | 0 | 144,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S consisting of `A`,`B`, and `C`.
Consider the (not necessarily contiguous) subsequences x of S that satisfy all of the following conditions:
* `A`, `B`, and `C` all occur the same number of times in x.
* No two adjacent characters in x are the same.
Among these subsequences, find one of the longest. Here a subsequence of S is a string obtained by deleting zero or more characters from S.
Constraints
* 1 \leq |S| \leq 10^6
* S consists of `A`,`B`, and `C`.
Input
Input is given from Standard Input in the following format:
S
Output
Print one longest subsequence that satisfies the conditions. If multiple solutions exist, any of them will be accepted.
Examples
Input
ABBCBCAB
Output
ACBCAB
Input
ABABABABACACACAC
Output
BABCAC
Input
ABCABACBCBABABACBCBCBCBCBCAB
Output
ACABACABABACBCBCBCBCA
Input
AAA
Output
Submitted Solution:
```
abc = "ABC"
acb = "ACB"
bac = "BAC"
bca = "BCA"
cab = "CAB"
cba = "CBA"
aabbcc = [abc, acb, bac, bca, cab, cba]
inds = []
notinds = []
S = str(input())
char = []
for i in aabbcc:
ind = S.find(i)
if ind > 0:
if ind not in inds:
inds.append(ind)
inds.append(ind+1)
inds.append(ind+2)
for i in range(len(S)):
if i not in inds:
notinds.append(i)
for i in S:
char.append(i)
notinds.sort()
notinds.reverse()
for i in notinds:
del char[i]
answer = "".join(char)
print(answer)
``` | instruction | 0 | 72,316 | 0 | 144,632 |
No | output | 1 | 72,316 | 0 | 144,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S consisting of `A`,`B`, and `C`.
Consider the (not necessarily contiguous) subsequences x of S that satisfy all of the following conditions:
* `A`, `B`, and `C` all occur the same number of times in x.
* No two adjacent characters in x are the same.
Among these subsequences, find one of the longest. Here a subsequence of S is a string obtained by deleting zero or more characters from S.
Constraints
* 1 \leq |S| \leq 10^6
* S consists of `A`,`B`, and `C`.
Input
Input is given from Standard Input in the following format:
S
Output
Print one longest subsequence that satisfies the conditions. If multiple solutions exist, any of them will be accepted.
Examples
Input
ABBCBCAB
Output
ACBCAB
Input
ABABABABACACACAC
Output
BABCAC
Input
ABCABACBCBABABACBCBCBCBCBCAB
Output
ACABACABABACBCBCBCBCA
Input
AAA
Output
Submitted Solution:
```
import collections
s = [s for s in input()]
t = s[0]
i = 1
while len(s) > i:
tmp = s
if t == s[i]:
tmp = s[:i]+s[i+1:]
t = s[i]
s = tmp
i+=1
c = collections.Counter(s)
m = min(c.values())
if m == 0:
print("")
exit()
mA = m
mB = m
mC = m
for i in range(len(s)):
if s[i] == "A":
if mA > 0:
print(s[i],end="")
mA-=1
elif s[i] == "B":
if mB > 0:
print(s[i],end="")
mB-=1
elif s[i] == "C":
if mC > 0:
print(s[i],end="")
mC-=1
print("")
``` | instruction | 0 | 72,317 | 0 | 144,634 |
No | output | 1 | 72,317 | 0 | 144,635 |
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