c++一般思路，为了便于理解同样采用双数组 原数组和差分数组

#include<iostream>
using namespace std;

constexpr int N = 1e+3 + 10;
int n,m,q;                                  // n行m列的矩阵 ， q为后面进行的操作次数
int a[N][N], b[N][N];                       // a为原数组   b为差分数组

void insert(int x1, int y1, int x2, int y2, int c)      // 更新差分数组的操作
{
    b[x1][y1] += c;
    b[x2 + 1][y1] -= c;
    b[x1][y2 + 1] -= c;
    b[x2+1][y2+1]  += c;
}

int main()
{
    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) insert(i,j,i,j, a[i][j]);


    while(q--)
    {
        int x1,y1,x2,y2,c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        insert(x1,y1,x2,y2,c);
    }

    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) a[i][j] = a[i][j-1] + a[i-1][j] - a[i-1][j-1] + b[i][j];     //求子矩阵的和

    for(int i = 1; i <= n; ++i)
    {
        for(int j = 1; j <= m; ++j)
        {
            cout << a[i][j] << " ";
        }
        cout << endl;   
    }

    return 0;
}

插入操作为啥这样写

更简洁的代码

#include<iostream>
using namespace std;

constexpr int N = 1e+3 + 10;
int n,m,q;
int a[N][N];

void insert(int x1, int y1, int x2, int y2, int c)
{
    a[x1][y1] += c;
    a[x2 + 1][y1] -= c;
    a[x1][y2 + 1] -= c;
    a[x2+1][y2+1]  += c;
}

int main()
{
    cin >> n >> m >> q;

    int d;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) cin >> d, insert(i,j,i,j,d);


    while(q--)
    {
        int x1,y1,x2,y2,c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        insert(x1,y1,x2,y2,c);
    }

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

    for(int i = 1; i <= n; ++i)
    {
        for(int j = 1; j <= m; ++j)
        {
            cout << a[i][j] << " ";
        }
        cout << endl;   
    }

    return 0;
}