codeforces 559B Equivalent Strings (最小表示法)

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:
a1 is equivalent to b1, and a2 is equivalent to b2
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.

Example
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”.

最小表示法，将s1折半深搜，然后判断左侧子串和右侧字串的关系，如果左侧子串大，交换左右字串，使得最后s1是按照规则可以得到最小的串。s2按照同样的方式转换，然后比较两个串

#include <bits/stdc++.h>
using namespace std;
const int maxn=200100;
char a[maxn],b[maxn];

int cmp(int len,char x[],char y[])
{
    for(int i=0;i<len;i++)
        if(x[i]<y[i]) return -1;
    else if(x[i]>y[i]) return 1;
    return 0;
}

void get(int len,char *s)
{
    if(len&1) return ;
    len/=2;
    get(len,s);
    get(len,s+len);
    if(cmp(len,s,s+len)>0)
    {
        for(int i=0;i<len;i++)
        {
            swap(s[i],s[i+len]);
        }
    }
}

int main()
{
    while(scanf(" %s %s",a,b)!=EOF)
    {
        get(strlen(a),a);
        get(strlen(b),b);
        if(strcmp(a,b)==0) printf("YES\n");
        else printf("NO\n");
    }
}