Codeforces548A:Mike and Fax

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".

题意：

给出一个字符串，判断是不是由k个等长回文串组成的

思路：

水题，暴力

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 500005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)

char str[1005];
int len;
int main()
{
    int i,j,k;
    while(~scanf("%s%d",str,&k))
    {
        len = strlen(str);
        if(len%k)
        {
            printf("NO\n");
            continue;
        }
        int r = len/k,flag = 0;
        for(i = 0; i<len; i+=r)
        {
            for(j=i; j<i+r; j++)
            {
                if(str[j]!=str[(i+r)-1-j+i])
                {
                    flag = 1;
                    break;
                }
            }
            if(flag)
                break;
        }
        if(flag)
            printf("NO\n");
        else
            printf("YES\n");
    }

    return 0;
}