Equalizing Two Strings CodeForces

大意: 给定两个串$s,t$, 每次操作任选长度$len$, 分别翻转$s,t$中一个长$len$的子串, 可以进行任意次操作, 求判断能否使$s$和$t$相同.

字符出现次数不一样显然无解, 否则若某种字符出现多次, 显然有解, 否则只要逆序对奇偶性相同就有解, 不同则无解.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(_i,_a,_n) for(int _i=_a;_i<=_n;++_i)
#define PER(_i,_a,_n) for(int _i=_n;_i>=_a;--_i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(_a) ({REP(_i,1,n) cout<<_a[_i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+10;
char s1[N],s2[N];
int n;

void work() {
	scanf("%d%s%s",&n,s1+1,s2+1);
	map<int,int> a1,a2;
	REP(i,1,n) ++a1[s1[i]],++a2[s2[i]];
	if (a1!=a2) return puts("NO"),void();
	for (auto &t:a1) if (t.y>=2) return puts("YES"),void();
	int f1=0,f2=0;
	REP(i,1,n) REP(j,1,i-1) f1+=s1[j]>s1[i],f2+=s2[j]>s2[i];
	puts((f1^f2)&1?"NO":"YES");
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) work();
}