codeforces 548A

A. Mike and Fax

time limit per test:

1 second

memory limit per test:

256 megabytes

input:

standard input

output

:standard output

While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.

He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.

He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.

Input

The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).

The second line contains integer k (1 ≤ k ≤ 1000).

Output

Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.

Sample test(s)

input

saba
2

output

NO

input

saddastavvat
2

output

YES

Note

Palindrome is a string reading the same forward and backward.

In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".

 

同样暴力，这题开始题目看错没看到每个回文串相等，浪费了不少时间。

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题解：

#include<cstdio>
#include<stdbool.h>
#include<string.h>
bool is[1001][1001];
char a[1001];
int n;
int lena;
bool flag;
int main()
{
    scanf("%s",a);
    scanf("%d",&n);
    lena=strlen(a);
    for(int i=1;i<=lena;i++)
    is[i][i]=1;
    for(int i=1;i<=lena-1;i++)
    if(a[i-1]==a[i])
    is[i][i+1]=1;
    for(int len=2;len<=lena-1;len++)
    for(int i=1;i+len<=lena;i++)
    {
        if(a[i-1]==a[i+len-1])
        is[i][i+len]=is[i+1][i+len-1];
    }
    int sum=0;
    if(lena%n)
    flag=0;
    else
    {
        for(int i=1;i<=n;i++)
        if(is[i*(lena/n)-(lena/n-1)][i*lena/n])
        sum++;
    }
    if(sum==n)
    flag=1;
    if(flag)
    printf("YES");
    else
    printf("NO");
}