P5300 [GXOI/GZOI2019]与或和

题目地址：P5300 [GXOI/GZOI2019]与或和

考虑按位计算贡献

对于 AND 运算，只有全 \(1\) 子矩阵才会有贡献

对于 OR 运算，所以非全 \(0\) 子矩阵均有贡献

如果求一个 01 矩阵中全 \(0/1\) 子矩阵的个数呢？

单调栈可以 \(O(n^2)\) 实现

总时间复杂度 \(O(n^2k)\) 其中 \(k\) 是二进制位数

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e3 + 6, P = 1e9 + 7;
int n, a[N][N], h[N], s[N], w[N], p;
ll ans[2];

inline ll get(int x) {
    return (1ll * x * (x + 1)) >> 1;
}

inline ll calc(int o) {
    ll cnt = 0;
    for (int j = 1; j <= n; j++) h[j] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if ((a[i][j] & 1) == o) ++h[j];
            else h[j] = 0;
            if (h[j] > s[p]) s[++p] = h[j], w[p] = 1;
            else {
                int k = 0;
                while (s[p] > h[j]) {
                    k += w[p];
                    cnt += (s[p] - max(s[p-1], h[j])) * get(k);
                    --p;
                }
                s[++p] = h[j], w[p] = k + 1;
            }
        }
        int k = 0;
        while (p) {
            k += w[p];
            cnt += (s[p] - s[p-1]) * (get(k));
            --p;
        }
    }
    return cnt % P;
}

void work(int o) {
    if (o == 32) return;
    ans[0] += calc(1) * (1 << o) % P, ans[0] %= P;
    ans[1] += (get(n) * get(n) % P - calc(0)) * (1 << o) % P, ans[1] %= P;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            a[i][j] >>= 1;
    work(o + 1);
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            scanf("%d", &a[i][j]);
    work(0);
    cout << (ans[0] + P) % P << " " << (ans[1] + P) % P << endl;
    return 0;
}