F. Equalizing Two Strings
You are given two strings s and t both of length n and both consisting of lowercase Latin letters.
In one move, you can choose any length len from 1 to n and perform the following operation:
Choose any contiguous substring of the string s of length len and reverse it;
at the same time choose any contiguous substring of the string t of length len and reverse it as well.
Note that during one move you reverse exactly one substring of the string s and exactly one substring of the string t.
Also note that borders of substrings you reverse in s and in t can be different, the only restriction is that you reverse the substrings of equal length. For example, if len=3 and n=5, you can reverse s[1…3] and t[3…5], s[2…4] and t[2…4], but not s[1…3] and t[1…2].
Your task is to say if it is possible to make strings s and t equal after some (possibly, empty) sequence of moves.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1≤q≤104) ― the number of test cases. Then q test cases follow.
The second line of the test case contains one string s consisting of n lowercase Latin letters.
The third line of the test case contains one string t consisting of n lowercase Latin letters.
Output
For each test case, print the answer on it ― "YES" (without quotes) if it is possible to make strings s and t equal after some (possibly, empty) sequence of moves and "NO" otherwise.
Example
input
4
4
abcd
abdc
5
ababa
baaba
4
asdf
asdg
4
abcd
badc
output
NO
YES
NO
YES
现在给你两个字符串，你可以进行若干次操作。
每次操作需要在每个字符串都选择出长度为len的一个区间，然后将这个区间的字符都进行翻转。
问你进行若干次操作后，这俩字符串能变成一样的吗？
按照这个顺序进行判断：
- 如果两个字符串存在不同的字符，那么肯定是NO
- 如果某个字符串存在两个相同的字符，那么一定是YES，因为可以就在这两个字符中进行无限次的翻转
- 如果两个字符串的逆的奇偶性相同，那么一定是YES
第三个怎么理解呢？在判断1和2之后，我们得到的一定是一个排列，问题就变成你可以翻转若干次，两个排列能否相同。
我们考虑我们同时翻转相同长度的，我们排列的逆一定会发生奇偶性的变化，那么如果一开始奇偶性就不同，那么不管怎么翻转，都不会相同。
#include<bits/stdc++.h> using namespace std; int n; string s1,s2,S1,S2; int Count(string s){ int num=0; for(int i=0;i<s.size();i++){ for(int j=0;j<i;j++){ if(s[j]>s[i]) num++; } } return num; } void solve(){ cin>>n>>S1>>S2; s1=S1;s2=S2; sort(s1.begin(),s1.end()); sort(s2.begin(),s2.end()); for(int i=0;i<n;i++){ if(s1[i]!=s2[i]){ puts("NO"); return; } } for(int i=1;i<n;i++){ if(s1[i]==s1[i-1]){ puts("YES"); return; } if(s2[i]==s2[i-1]){ puts("YES"); return; } } if(Count(S1)%2==Count(S2)%2){ puts("YES"); }else{ puts("NO"); } return; } int main(){ int t; scanf("%d",&t); while(t--)solve(); } 来源：博客园
作者：qscqesze
链接：https://www.cnblogs.com/qscqesze/p/11799050.html