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.

$\text{[math]}$

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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int main() {
 5     string str;
 6     int k,n;
 7     int id;
 8     cin>>str>>k;
 9     /**如果恰没有k个的时候是输出NO 没考虑这个条件一直wa*/
10     if(str.length()%k != 0) {
11         cout<<"NO"<<endl;
12         return 0;
13     } 
14     n = str.length()/k;
15     bool ans = false;
16     /**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*/ 
17     for(int i = 1; i <= k; i++) {
18         int bi = (i-1)*n;
19         int ei =  i*n-1;
20         while(bi <= ei) {
21             if(str[bi] != str[ei]) {
22                 ans = true;
23                 break;
24             }
25             ei--;
26             bi++;
27         }
28         if(ans) break;
29     }
30     if(ans) cout<<"NO"<<endl;
31     else cout<<"YES"<<endl;
32     return 0;
33 }