Codeforces Round #305 (Div. 2) A. Mike and Fax 暴力回文串

                                     A. Mike and Fax

                                                               Time Limit: 20 Sec  Memory Limit: 256 MB

 

题目连接

http://codeforces.com/contest/548/problem/A

 

Description

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 Input

saba
2

Sample Output

NO

 

 

题解：

题目意思就是给你一个字符串，判断由K个长度相同的回文串组成，是否正确

直接暴力判断就好了  

 

 

//llm
#include<stdio.h>
#include<string.h>
#include<math.h>
#define N 1005
char str[N];
int main()
{
    int l,i,j,k,n,t,len,flag;
    while(scanf("%s",str)!=EOF)
    {
        flag=0;
        scanf("%d",&k);
        getchar();
        len=strlen(str);
        n=len/k;
        //printf("%c %c\n",str[0],str[len-1]);
        if(k*n!=len) flag=1;
        else 
        {
            for(i=0;i<len;i+=n)
            {
                l=i;
                j=i;
                while(i<j+n-1)
                {
                    if(str[l]==str[j+n-1])
                    {
                        l++;
                        j--;
                    }
                    else 
                    {
                        flag=1;
                        break;
                    }
                }
            }
        }
        if(flag==1) printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}