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.Example Input
saba 2 Output NO Input saddastavvat 2 Output YESNote
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". #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; char ma[2322]; int main() { int n; while(~scanf("%s %d", ma, &n)) { int len = strlen(ma); if(len%n)//回文串 { printf("No\n"); continue; } int r = len / n;//每个的长度 int flag = 0;//标记 for(int i=0;i<len;i+=r) { for(int j=i;j<i+r;j++) { if(ma[j]!=ma[(i+r)-1-j+i]) { flag = 1; break; } } if(flag) break; } if(flag) printf("NO\n"); else printf("YES\n"); } return 0; }