Codeforces 560D Equivalent Strings【Dfs】

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本文介绍了一种新的字符串等价性定义及判断方法，通过将字符串分为两部分并比较其等价性来解决相关问题。提供了AC代码实现。

Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent 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.

Examples

Input

aaba
abaa

Output

YES

Input

aabb
abab

Output

NO

Note

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".

题目大意：

给你两个长度相等的字符串，让你判断两个字符串是否相等。两个字符串相等的条件有两个，只要满足一个条件就算是相等的：

一、当前字符串是相等的。

二、将其分成前后两段长度相等的子串之后，a1==b1&&a2==b2||a1==b2&&a2==b1。

思路：

1、我们直接按照题意模拟即可。

2、注意一些常数问题，写的渣可能会TLE掉。注意一些细节，别的就没什么可说的了。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char a[200500];
char b[200500];
char aa[200500];
char bb[200500];
int vis[350];
int n,ok;
int judge(int l1,int r1,int l2,int r2)
{
    memset(vis,0,sizeof(vis));
    for(int i=l1; i<=r1; i++)
    {
        vis[a[i]]++;
    }
    for(int i=l2; i<=r2; i++)
    {
        vis[b[i]]--;
    }
    for(int i=1; i<=335; i++)
    {
        if(vis[i]!=0)return 0;
    }
    return 1;
}
int Dfs(int l1,int r1,int l2,int r2,int len)
{
    if(judge(l1,r1,l2,r2)==0)return 0;
    if(l1==r1&&l2==r2)
    {
        if(a[l1]==b[l2])
        return 1;
        else return 0;
    }
    int flag=1;
    int j=l2;
    for(int i=l1;i<=r1;i++)
    {
        if(a[i]==b[j])j++;
        else flag=0;
    }
    if(flag==1)return 1;
    if(len%2==1)
    {
        return 0;
    }
    if((Dfs(l1,(l1+r1)/2,l2,(l2+r2)/2,len/2)&&Dfs((l1+r1)/2+1,r1,(l2+r2)/2+1,r2,len/2)))
    {
        return 1;
    }
    if(Dfs(l1,(l1+r1)/2,(l2+r2)/2+1,r2,len/2)&&Dfs((l1+r1)/2+1,r1,l2,(l2+r2)/2,len/2))
    {
        return 1;
    }
    return 0;
}
int main()
{
    while(~scanf("%s%s",a,b))
    {
        memset(vis,0,sizeof(vis));
        n=strlen(a);
        if(judge(0,n-1,0,n-1))
        {
            int ok;
            ok=Dfs(0,n-1,0,n-1,n);
            if(ok==1)printf("YES\n");
            else printf("NO\n");
        }
        else printf("NO\n");
    }
}