【开源社区openEuler实践】D2. Burenka and Traditions (hard version)
【代码】Codeforces Round 814 (Div. 2) D2. Burenka and Traditions (hard version)
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思路:

#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e9, maxm = 4e4 + 5;
const int N = 1e5;
const int mod = 1e9 + 7;
// const int mod = 998244353;
//const __int128 mod = 212370440130137957LL;
// int a[505][5005];
// bool vis[505][505];
int n, m;
int a[maxn];
string s;
struct Node{
// int val, id;
// bool operator<(const Node &u)const{
// return val < u.val;//!!!!!!!!!!!!!!!!!!
// }
int v, w;
};
//long long ? maxn ? n? m?
void solve(){
int res = 0;
int k, q;
cin >> n;
int sum = 0;
for(int i = 1; i <= n; i++){
cin >> a[i];
}
map<int, int> mp;
int Xor = 0;
res = n;
mp[0] = 1;//!!!!!!!!
for(int i = 1; i <= n; i++){
Xor ^= a[i];
if(mp[Xor]){//两个相同的前缀异或和可以确定一个异或和为0的子区间
res--;
mp.clear();//保证找出的异或和为0的子区间不重叠
}
mp[Xor] = 1;
}
cout << res << '\n';
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
int T = 1;
cin >> T;
while (T--)
{
solve();
}
return 0;
}
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