3thweek——E.暴力求解+DFS

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

案例输入输出：

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".

分析步骤：

1.先判断两串字符串长度为奇数还是偶数，奇数则要遍历全部的两个字符串，字符和顺序都相同才能输出YES，否则输出NO。

2.如果是偶数，两个字符串完全相同就直接输出YES，否则将两个字符串各切成两个子字符串再比较。

代码如下：

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=200005;
char a[maxn],b[maxn];

int i,flag=0;
int compare(char* a1,char* b1,int l)  //自定义比较函数
{   
    for(i=0;i<l;i++)
    {
        if(a1[i]!=b1[i])
        {
            return 0;
        }
    }
    flag=1;
    return flag;
}
    

int fun(char* a1,char* b1,int l)
{   
    int k=0;
    if(compare(a1,b1,l))
           k=1; 
    if(l%2==1)
    {
        if(k==1)
           return 1;
        else
            return 0;
    }    
    else
    {
        if(k==1)
            return 1;
    else
    {
            l=l/2;
          if( (fun(a1,b1,l)  && fun(a1+l,b1+l,l)) || (fun(a1,b1+l,l) && fun(a1+l,b1,l)) )   
            return 1;
          else
              return 0;
    }
    }
}

int main()
{
    cin>>a;
    cin>>b;
    int len=strlen(a);
    if(fun(a,b,len))
       cout<<"YES"<<endl;
    else
        cout<<"NO"<<endl;
return 0;
}