Equivalent Strings (equal Strings ?), Equivalentstrings
Equivalent Strings E-brute force solution, DFSTime Limit:2000 MSMemory Limit:262144KB64bit IO Format:% I64d & % I64u
Description
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two stringsAAndBOf equal length are calledequivalent in one of the two cases:
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 limit 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
Input
aaba
abaa
Output
YES
Input
aabb
abab
Output
NO
Hint
In the first sample you shoshould 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" + "".
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 ".
Give two strings to determine whether they are equal. If they are equal, one is that the strings are directly equal, one is to cut the two parts with the same length and then the two substrings are equal (two cases)
Problem: the row is determined directly. If the current length is an odd number, if not completely equal, 0 is returned directly. Otherwise, two cases are considered.
# Include <stdio. h> # include <string. h> char a [200005], B [200005]; int juge (char * p, char * q, int len) {if (! Strncmp (p, q, len) // determines whether the string length of p, q is equal to return-1; if (len % 2) return 0; // end judgment int mid = len/2; if (juge (p, q + mid, mid) & juge (p + mid, q, mid) return-1; if (juge (p + mid, q + mid, mid) & juge (p, q, mid) return-1 ;}int main () {scanf ("% s", & a, & B); if (juge (a, B, strlen (a) printf ("YES "); else printf ("NO"); return 0 ;}