【开源社区openEuler实践】D2. Burenka and Traditions (hard version)

原创已于 2025-01-03 15:57:49 修改 · 203 阅读

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#算法

于 2024-04-18 00:32:10 首次发布

题目

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;
}