ACM判断字符串”相等“

Description

Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are calledequivalent in one of the two cases:

1. They are equal.
2. If we split string a into two halves of the same size a₁ and a₂, and string b into two halves of the same size b₁ and b₂, then one of the following is correct:
   1. a₁ is equivalent to b₁, and a₂ is equivalent to b₂
   2. a₁ is equivalent to b₂, and a₂ is equivalent to b₁

As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.

Gerald has already completed this home task. Now it's your turn!

Input

The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.

Output

Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.

Sample Input

Input

aaba
abaa

Output

YES

Input

aabb
abab

Output

NO

Hint

In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".

In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".

解题思路：

题目的大意是给定两个字符串，判断这两个字符串是不是“相等”的。“相等”的条件是a b两个字符串被拆成a1,a2,b1,b2时，a1=b1并且a2=b2或者a1=b2并且a2=b1这样时a b两个字符串是"相等"的。所以我们可以先判断这两个字符串是不是相等的，如果是的话，那这两个字符串就是“相等的”，还有如果字符串的长度是奇数，那么就是肯定不想等的了。不断的进行递归判断。

程序代码：

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
const int m = 200010;
char s1[m], s2[m];
bool dfs(char *a,char *b,int n)
{
   if (strncmp(a, b, n) == 0)              //比较字符串
       return true;
   if (n % 2 != 0)                          //判断字符串长度为偶数还是奇数
       return false;
   int len = n / 2;                        //平分字符串
   
   if (dfs(a, b + len, len) && dfs(a + len, b, len))
       return true;                                    
   if (dfs(a, b, len) && dfs(a + len, b + len, len))    //拆分后的字符串比较
       return true;
   return false;
}
int main()
{
    cin >> s1;
    cin >> s2;
    cout <<( dfs(s1, s2, strlen(s1)) ? "YES" : "NO") << endl;   
    return 0;