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 kpalindromes 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

 

HINT

 

题意

给你个字符串，告诉你这个字符串是由k个相同长度的回文串构成的，判断是否正确

题解：

 数据范围很小，暴力判断就好了

#include<bits/stdc++.h>
using namespace std;

int main() {
    string str;
    int k,n;
    int id;
    cin>>str>>k;
    /**如果恰没有k个的时候是输出NO 没考虑这个条件一直wa*/
    if(str.length()%k != 0) {
        cout<<"NO"<<endl;
        return 0;
    } 
    n = str.length()/k;
    bool ans = false;
    /**分成K组进行判断 如果有一组不满足回文串条件  则确定输出NO*/ 
    for(int i = 1; i <= k; i++) {
        int bi = (i-1)*n;
        int ei =  i*n-1;
        while(bi <= ei) {
            if(str[bi] != str[ei]) {
                ans = true;
                break;
            }
            ei--;
            bi++;
        }
        if(ans) break;
    }
    if(ans) cout<<"NO"<<endl;
    else cout<<"YES"<<endl;
    return 0;
}