Codeforces Round #313 (Div. 1) B. Equivalent Strings DFS暴力

B. Equivalent Strings

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/559/problem/B

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

aaba
abaa

Sample Output

YES

HINT

题意

判断俩字符串是否相似，相似的条件如下：

a1+a2=A,a1和a2都是A的一半

b1+b2=B,同理

如果A,B相等，那么相似

如果A的长度为偶数{

　　如果a1与b1,a2与b2相似或者a1与b2，b1与a2相似

　　那么A,B相似

}

否则不相似

题解：

就如同题目讲的一样，傻逼暴力DFS就好了

不要想多了

代码

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=;
char A[MAXN],B[MAXN];
bool cmp(char x[],char y[],int len)
{
    //printf("%d %d\n",x-A,y-B);
    bool isok=;
    for(int i=;i<len;i++)
        if(x[i]!=y[i])isok=;
    return isok;
}
bool equ(char x[],char y[],int len)
{
    //printf("%d %d %d\n",x-A,y-B,len);
    if(cmp(x,y,len))return ;
    if(len%==)
        return (equ(x,y+len/,len/)&&equ(x+len/,y,len/))
                ||(equ(x,y,len/)&&equ(x+len/,y+len/,len/));
    return ;
}
int main()
{
    scanf("%s%s",A,B);
    int len=strlen(A);
    printf("%s\n",equ(A,B,len) ? "YES" : "NO");
    return ;
}