CodeForces - 548A (字符串)

最新推荐文章于 2018-08-29 09:29:55 发布

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本文介绍了一个算法问题：如何判断给定字符串是否由相同长度的k个回文串拼接而成。通过分析输入字符串和指定数量k，文章提供了一种有效的解决方案，并附带示例进行说明。

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

YES

Note

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;
}