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Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

# Codeforces Round #598 (Div. 3) F. Equalizing Two Strings 构造

## 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 first line of the test case contains one integer n (1≤n≤2⋅105) — the length of s and t.

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.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 (∑n≤2⋅105).

## 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.

input
4
4
abcd
abdc
5
ababa
baaba
4
asdf
asdg
4
abcd
output
NO
YES
NO
YES

## 题解

1. 如果两个字符串存在不同的字符，那么肯定是NO
2. 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
3. 如果两个字符串的逆的奇偶性相同，那么一定是YES

## 代码

#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();
}
posted @ 2019-11-05 15:58  qscqesze  阅读(295)  评论(1编辑  收藏