codeforces(559B)--B. Equivalent Strings(暴搜 或 最小表示法)

B. Equivalent Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

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".

定义相等的字符串为它的等分成的两个字串相等，或交叉相等。问给出的两个字串是不是同样

直接暴搜，，。，。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
char s1[210000] , s2[210000] ;
int solve(int l1,int r1,int l2,int r2) {
    int i , mid1 , mid2 ;
    for(i = 0 ; i <= (r1-l1) ; i++)
        if( s1[l1+i] != s2[l2+i] ) break ;
    if( i > (r1-l1) ) return 1 ;
    if( (r1-l1+1)%2 ) return 0 ;
    mid1 = (l1+r1)/2 ;
    mid2 = (r2+l2)/2 ;
    if( solve(l1,mid1,l2,mid2) && solve(mid1+1,r1,mid2+1,r2) ) return 1 ;
    if( solve(l1,mid1,mid2+1,r2) && solve(mid1+1,r1,l2,mid2) ) return 1 ;
    return 0 ;
}
int main() {
    int l1 , l2 ;
    while( scanf("%s %s", s1, s2) != EOF ) {
        l1 = strlen(s1) ;
        l2 = strlen(s2) ;
        if( l1 != l2 || !solve(0,l1-1,0,l2-1) )
            printf("NO
") ;
        else
            printf("YES
") ;
    }
    return 0 ;
}

最小表示法，将s1折半深搜，然后推断左側子串和右側字串的关系，假设左側子串大。交换左右字串，使得最后s1是依照规则能够得到最小的串。

s2依照相同的方式转换，然后比較两个串

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
char s1[210000] , s2[210000] ;
char c ;
void solve(int l,int r,char *s) {
    if( (r-l+1)%2 ) return ;
    int i , mid = (l+r)/2 ;
    solve(l,mid,s) ;
    solve(mid+1,r,s) ;
    for(i = 0 ; i < (r-l+1)/2 ; i++) {
        if( s[l+i] > s[mid+1+i] ) break ;
        if( s[l+i] < s[mid+1+i] ) return ;
    }
    if( i == (r-l+1)/2 ) return ;
    for(i = 0 ; i < (r-l+1)/2 ; i++) {
        c = s[l+i] ; s[l+i] = s[mid+1+i] ; s[mid+1+i] = c ;
    }
}
int main() {
    int l1 , l2  ;
    while( scanf("%s %s", s1, s2) != EOF ) {
        l1 = strlen(s1) ;
        l2 = strlen(s2) ;
        solve(0,l1-1,s1) ;
        solve(0,l2-1,s2) ;
        if( l1 != l2 || strcmp(s1,s2) )
            printf("NO
") ;
        else
            printf("YES
") ;
    }
    return 0 ;
}