二维前缀和 & 二维差分

二维前缀和

求前缀和：(s_{i, : j} = s_{i - 1, : j} + s_{i, : j - 1} - s_{i - 1, : j - 1} + a_{i, : j})

算部分和：(s_{x_2, : y_2} - s_{x_1 - 1, : y_2} - s_{x_2, : y_1 - 1} + s_{x_1 - 1, : y_1 - 1})

#include <bits/stdc++.h>
using namespace std;

const char nl = '
';
const int N = 1000 + 50;

int a[N][N], s[N][N];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m, q;
    cin >> n >> m >> q;

    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            cin >> a[i][j];

    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];

    while (q--){
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        cout << s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1] << nl;
    }

    return 0;
}

二维差分

(b_{x_1, : y_1} : += : c)

(b_{x_2 + 1, : y_1} : -= : c)

(b_{x_1, : y_2 + 1} : -= : c)

(b_{x_2 + 1, : y_2 + 1} : += : c)

#include <bits/stdc++.h>
using namespace std;

const char nl = '
';
const int N = 100 + 50;

int a[N][N], b[N][N];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m, q;
    cin >> n >> m >> q;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            cin >> a[i][j];

    // 构造
    for (int i = 1; i <= n; ++i){
        for (int j = 1; j <= m; ++j){
            b[i][j] += a[i][j];
            b[i + 1][j] -= a[i][j];
            b[i][j + 1] -= a[i][j];
            b[i + 1][j + 1] += a[i][j];
        }
    }

    // 区间操作
    while (q--){
        int x1, y1, x2, y2, c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        b[x1][y1] += c;
        b[x2 + 1][y1] -= c;
        b[x1][y2 + 1] -= c;
        b[x2 + 1][y2 + 1] += c;
    }

    // 计算前缀和
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j)
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];

    // 输出
    for (int i = 1; i <= n; ++i){
        for (int j = 1; j <= m; ++j) cout << b[i][j] << ' ';
        cout << nl;
    }

    return 0;
}