2017-4-21-Train:Codeforces Round #305 (Div. 2)

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A. Mike and Fax(模拟 + two points)

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.

Examples

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".

Means:

求给定的字符串中是否有K个回文子串

Solve:

模拟题,先判断能否整除k,如果能,长度就是length / k,然后two points扫一遍

Code:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 static const int MAXN = 1e3 + 10;
 4 int k;
 5 char data[MAXN];
 6 int main(int argc, char const *argv[])
 7 {
 8     //freopen("D:\\系统优化\\Desktop\\littlepea\\in.data" , "r" , stdin);
 9     scanf(" %s" , data + 1);
10     int len = strlen(data + 1);
11     scanf("%d" , &k);
12     if(len % k || k > len)
13     {
14         puts("NO");
15         return 0;
16     }
17     int cut = len / k;
18     bool flag = 1;
19     for(int i = 1 ; i <= len ; i += cut)
20     {
21         int l = i , r = i + cut - 1;
22         while(l < r)
23         {
24             if(data[l] != data[r])
25             {
26                 flag = 0;
27                 break;
28             }
29             ++l , --r;
30         }
31         if(!flag)
32             break;
33     }
34     if(!flag)
35         puts("NO");
36     else
37         puts("YES");
38     return 0;
39 }

View Code

2017-4-21-Train:Codeforces Round #305 (Div. 2)

标签：put   技术   read   return   字符   rem   ges   break   closed   

原文地址：http://www.cnblogs.com/jianglingxin/p/6745552.html