A - Mike and Fax

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Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

 

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

 

Input

saba
2

Output

NO

Input

saddastavvat
2

Output

YES

Hint

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".

判断等长回文串个数。

附AC代码：

 1 #include<iostream>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 int main(){
 7     string s;
 8     int n;
 9     cin>>s;
10     cin>>n;
11     int flag=0;
12     if(s.size()%n){
13         cout<<"NO"<<endl;
14     }
15     else{
16         int len=s.size();
17         int t=len/n;
18         for(int i=0;i<len;i+=t){
19             for(int j=i;j<(i+t/2);j++){
20                 if(s[j]!=s[i+t-1-j+i]){
21                     flag=1;
22                     break;
23                 }
24             }
25         } 
26         if(flag){
27             cout<<"NO"<<endl;
28         }
29         else{
30             cout<<"YES"<<endl;
31         }
32     }
33     return 0;
34 }

A - Mike and Fax

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原文地址：http://www.cnblogs.com/Kiven5197/p/5720389.html