Equivalent Strings （字符串相等？）

 Equivalent Strings 

 

E - 暴力求解、DFS

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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".

题意：给两个字符串，判断它们是否相等，相等有两种情况，一个是字符串直接相等，一个是切成长度相同的两份以后两子串相等（两种情况）

题解：直接判断行了，如果当前长度为奇数，如果不是完全相等，直接返回0，否则分两种情况判断

#include<stdio.h>
#include<string.h>
char a[200005],b[200005];

int juge(char *p,char *q, int len)
{
    if(!strncmp(p,q,len))        //判断怕，p，q字符串长度是否相等
        return -1;
    if(len%2)
        return 0;                     // 结束判断
    int mid=len/2;
    if(juge(p,q+mid,mid)&&juge(p+mid,q,mid))
        return -1;
    if(juge(p+mid,q+mid,mid)&&juge(p,q,mid))
        return -1;
}

int main()
{
    scanf("%s%s",&a,&b);
    if(juge(a,b,strlen(a)))
        printf("YES");
    else
        printf("NO");
    return 0;
}