Equivalent Strings

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:

They are equal.
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".

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef unsigned long long LL;
 4 const int maxn=200010;
 5 const int B=123;
 6 char a[maxn],b[maxn];
 7 int cmp(int len,char x[],char y[])
 8 {
 9     for(int i=0;i<len;i++)
10         if(x[i]<y[i])return -1;
11         else if(x[i]>y[i])return 1;
12     return 0;
13 }
14 void get(int len,char *s)
15 {
16     if(len&1)return ;
17     len/=2;
18     get(len,s);
19     get(len,s+len);
20     if(cmp(len,s,s+len)>0)
21     {
22         for(int i=0;i<len;i++)
23             swap(s[i],s[i+len]);
24     }
25 }
26 int main()
27 {
28     while(scanf("%s%s",a,b)!=EOF)
29     {
30         get(strlen(a),a);
31         get(strlen(b),b);
32         if(strcmp(a,b)==0)printf("YES
");
33         else printf("NO
");
34     }
35     return 0;
36 }

View Code

一开始我也是在傻傻的暴力，然而并没有什么软用。。

仔细想想就知道会超时。。

其实我们只要按他的string比较规则把字符串一层层比较然后排序，

在最后return string a==string b 就行了

这样是不是剪短了简答树的很多子叶呢？？