Codeforces Round #313 D. Equivalent Strings（DFS）

D. Equivalent Strings

time limit per test

memory limit per test

input

output

a and b of equal length are calledequivalent

1. They are equal.
2. 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

1 to 200 000

Output

YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.

Sample test(s)

input

aaba abaa

output

YES

input

aabb abab

output

NO

Note

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".

aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".

#include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>

    using namespace std;

    char a[200010],b[200010];

    int DFS(char *a,char *b,int l)
    {
        if(strncmp(a,b,l) == 0)
        {
            return 1;
        }
        if(l%2)
        {
            return 0;
        }
        int p = l / 2;
        if((DFS(a,b+p,p) && DFS(a+p,b,p)) || (DFS(a,b,p) && DFS(a+p,b+p,p)))
        {
            return 1;
        }
        return 0;
    }

    int main()
    {
        while(scanf("%s",a)!=EOF)
        {
            scanf("%s",b);
            int len = strlen(a);
            int pk = DFS(a,b,len);
            if(pk == 1)
            {
                printf("YES\n");
            }
            else
            {
                printf("NO\n");
            }
        }
        return 0;
    }