[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： \(1\leq n\leq 10^5, a_i,k\leq 10^5\)

CF 数据范围： \(1\leq n\leq 10^5, a_i,k\leq 10^6\)

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\n", ans[i]);

}

int main() {work(); return 0; }