2021杭电多校赛第一场

Xor sum

#include <bits/stdc++.h>
using namespace std;
class Trie {
private:
    static const int MAXN = 1e5 + 5;
    static const int BASE = 30;
    int idx = 1, trie[MAXN * BASE][2];
    int pos[MAXN * BASE]; // pos[i]: 异或前缀为i的最右端点
public:
    void insert(int p, int x) {
        int u = 1;
        for (int i = BASE; i >= 0; --i) {
            int v = (x >> i) & 1;
            if (!trie[u][v]) trie[u][v] = ++idx;
            u = trie[u][v];
            pos[u] = max(pos[u], p);
        }
    }
    int find(int x, int k) {
        int u = 1, res = -1;
        for (int i = BASE; i >= 0; --i) {
            int v = (x >> i) & 1;
            if ((k >> i) & 1) { // 如果k当前位为1
                                // 只有当前位x能找和它异或为1的数才有可能大于等于k
                                // 否则直接返回已经找到的最大值
                if (trie[u][v ^ 1]) u = trie[u][v ^ 1];
                else return res;
            } else { // 如果k当前位为0, x异或后当前位是0是1都有可能大于等于k
                     // 如果异或后是1, 那肯定大于等于k了, 不用往下再遍历了, 直接尝试更新最大值
                     // 否则继续往下遍历
                if (trie[u][v ^ 1]) res = max(res, pos[trie[u][v ^ 1]]);
                u = trie[u][v];
            }
        }
        res = max(res, pos[u]); // 到达最底层即==k的情况
        return res;
    }
    void clear() {
        for (int i = 1; i <= idx; ++i) {
            pos[i] = 0;
            trie[i][0] = trie[i][1] = 0;
        }
        idx = 1;
    }
} tree;
const int MAXN = 1e5 + 5;
int a[MAXN];
int main(int argc, char *argv[]) {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int T;
    cin >> T;
    while (T--) {
        tree.clear();
        int n, k;
        cin >> n >> k;
        for (int i = 1; i <= n; ++i) {
            cin >> a[i];
            a[i] ^= a[i - 1];
        }
        int L = -1, R = n + 1;
        for (int r = 1; r <= n; ++r) {
            tree.insert(r, a[r]);
            int l = tree.find(a[r], k);
            if (~l && r - l < R - L) {
                L = l, R = r;
            }
        }
        if (L == -1) cout << -1 << '\n';
        else cout << L + 1 << ' ' << R << '\n';
    }
    system("pause");
    return 0;
}