[CQOI 2018]异或序列&[Codeforces 617E]XOR and Favorite Number

Description

题库链接1 题库链接2

已知一个长度为 (n) 的整数数列 (a_1,a_2,cdots,a_n) ，给定查询参数 (l,r) ，问在 ([l,r]) 区间内，有多少连续子序列满足异或和等于 (k) 。

CQOI 数据范围： (1leq nleq 10^5, a_i,kleq 10^5)

CF 数据范围： (1leq nleq 10^5, a_i,kleq 10^6)

Solution

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Code

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+5, SIZE = 1<<20;

int n, m, k, lim, a[N], cnt[SIZE]; ll ans[N];
struct tt {
    int l, r, id;
    bool operator < (const tt &b) const {
        return l/lim == b.l/lim ? r < b.r : l < b.l;
    }
}b[N];

void work() {
    scanf("%d%d%d", &n, &m, &k); lim = sqrt(n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] ^= a[i-1];
    for (int i = 1; i <= m; i++) scanf("%d%d", &b[i].l, &b[i].r), --b[i].l, b[i].id = i;
    sort(b+1, b+m+1);
    int curl = 1, curr = 0; ll sum = 0;
    for (int i = 1; i <= m; i++) {
        int l = b[i].l, r = b[i].r;
        while (curr < r) sum += cnt[a[++curr]^k], ++cnt[a[curr]];
        while (curl > l) sum += cnt[a[--curl]^k], ++cnt[a[curl]];
        while (curr > r) --cnt[a[curr]], sum -= cnt[a[curr--]^k];
        while (curl < l) --cnt[a[curl]], sum -= cnt[a[curl++]^k];
        ans[b[i].id] = sum;
    }
    for (int i = 1; i <= m; i++) printf("%lld
", ans[i]);
}
int main() {work(); return 0; }