CF 559B Equivalent Strings 分治05 A题

训练
原题

题目描述
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.

Examples
inputCopy
aaba
abaa
outputCopy
YES
inputCopy
aabb
abab
outputCopy
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”.

题意

给定两个字符串判等的标准（are called equivalent），即两个串 are equivalent ，或者各自对半分成两个字串，交叉are equivalent满足至少一个条件，则说明两串 are equivalent。
之所以飙英文，是因为直接说“是相等的“会晕，这里的equivalent应理解为等价

思考过程

开始有点困，竟然还以为只用分一次
然后觉得不对劲，但是又把样例的aaba看成了aaab（反正就是看成两个ab），然后迷迷糊糊敲了一堆，才发现又不对劲。。
然后思路出来了，
用4个参数表示两个需要判断等价的字串的首位index，然后描述等价条件分治到底

测奇数长度，莫名其妙会l>r然后一顿乱加条件
然后就wa第22点
最后调的时候才注意到，要分成“two halves of the same size“！！！！

#include <string.h>
#include <cstring>
#include <iostream>
//#include<ctring>

using namespace std;
string s1, s2;
bool eql(int l, int r, int ll, int rr) {
  if (l > r) return 0;
  if (l == r && (ll == rr)) return s1[l] == s2[ll];
  int t = 0,x=0;
  int ls = r - l + 1;
  for (register int i = 0; i <= ls - 1; i++,x=i) {//判断两个串是否完全相同，如果是奇数，不能分成两段相同长度的串，但如果完全相等也合法
    if (s1[i + l] != s2[i + ll]) break;
    t = i;//如果加到i++后面，就会跟着i在最后一次循环后+1！！！！！
  }
  if (t == ls - 1) return 1;
  
  if (ls % 2) return 0;

  int mid1 = (l + r) / 2;

  if (eql(l, mid1, ll, mid2) && eql(mid1 + 1, r, mid2 + 1, rr)) return 1;//对应eql
  if (eql(l, mid1, mid2 + 1, rr) && eql(mid1 + 1, r, ll, mid2)) return 1;//交叉eql

  return 0;
}
int main() {
  cin >> s1 >> s2;
  int l1 = s1.length(), l2 = s2.length();
  s1 = " " + s1, s2 = " " + s2;
  if (eql(1, l1, 1, l1))
    cout << "YES";
  else
    cout << "NO";
}