1872E. Data Structures Fan · znzryb

## 思路讲解

哈哈，其实说起来也简单，在xor中，加上这个数和减去这个数都是xor，因此对区间进行反转操作，就是对这个区间进行异或就可以了，不用区分是加上还是减去。

当然，怎么对区间进行异或那？其实也很简单，异或前缀和。

## AC代码

[https://codeforces.com/problemset/submission/1872/309601127](https://codeforces.com/problemset/submission/1872/309601127)

```cpp
// Problem: E. Data Structures Fan
// Contest: Codeforces - Codeforces Round 895 (Div. 3)
// URL: https://codeforces.com/problemset/problem/1872/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(1e5)+10,INF=static_cast<ll>(5e18)+3;

ll N,T,A[MAXN];
char S[MAXN];
ll sum[MAXN];
inline void solve(){
	cin>>N;
	sum[0]=0;
	for(int i=1;i<=N;++i){
		cin>>A[i];
		sum[i]=sum[i-1]^A[i];
	}
	ll res1=0,res0=0;
	for(int i=1;i<=N;++i){
		cin>>S[i];
		if(S[i]=='1'){
			res1^=A[i];
		}else{
			res0^=A[i];
		}
	}
	ll Q;
	cin>>Q;
	for(int i=1;i<=Q;++i){
		ll op;
		cin>>op;
		if(op==1){
			ll l,r;
			cin>>l>>r;
			res1^=(sum[r]^sum[l-1]);
			res0^=(sum[r]^sum[l-1]);
		}else{
			ll ind;
			cin>>ind;
			if(ind){
				cout<<res1<<" ";
			}else{
				cout<<res0<<" ";
			}
		}
	}
	cout<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	return 0;
}
/*
AC
https://codeforces.com/problemset/submission/1872/309601127
*/
```

## 心路历程（WA，TLE，MLE……）