#include <iostream> 
using namespace std;
#define ll long long
ll n, m, k, x1, y1, x2, y2, s,a[1001][1001], sum[1001][1001];
int main()
{
    ios::sync_with_stdio(0);
    cin >> n >> m >> k;
    for (ll i = 1; i <= n; i++)
    {
        for (ll j = 1; j <= m; j++)
        {
            cin >> a[i][j];
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + a[i][j];
        }
    }
    for (ll i = 1; i <= k; i++)
    {
        cin >> x1 >> y1 >> x2 >> y2;
        s = sum[x2][y2] - (sum[x1 - 1][y2] + sum[x2][y1 - 1] - sum[x1 - 1][y1 - 1]);
        cout << s << endl;
    }
    return 0;
}

因为y1所以只能用iostream了

#include <bits/stdc++.h>
using namespace std;
const int MAX=1010;
int n,m,k,cnt[MAX][MAX],prefix[MAX][MAX];
int main(){
	scanf("%d%d%d",&n,&m,&k);
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			scanf("%d",&cnt[i][j]);
			prefix[i][j]=prefix[i-1][j]+prefix[i][j-1]-prefix[i-1][j-1]+cnt[i][j];
		} 
	}
	int l1,l2,r1,r2;
	for (int i=1;i<=k;i++){
		scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
		printf("%d\n",prefix[l2][r2]-prefix[l1-1][r2]-prefix[l2][r1-1]+prefix[l1-1][r1-1]);
	}
	return 0;
}

#include <bits/stdc++.h> 
using namespace std;
int main()
{
	int n, m, k, x1, y1, x2, y2, s;
    cin >> n >> m >> k;
    int a[n + 1][m + 1], sum[n + 1][m + 1];
    memset(sum, 0, sizeof(sum));
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> a[i][j];
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + a[i][j];
        }
    }
    for (int i = 1; i <= k; i++)
    {
        cin >> x1 >> y1 >> x2 >> y2;
        s = sum[x2][y2] - (sum[x1 - 1][y2] + sum[x2][y1 - 1] - sum[x1 - 1][y1 - 1]);
        cout << s << endl;
    }
    return 0;
}

此题是较简单的二维前缀和模板题 前缀和、差分复习直播 链接：https://www.bilibili.com/video/BV1MN4y1j7dR?spm_id_from=333.999.0.0&vd_source=3136e9894179057b25e6f1820e1738da

代码如下：

#include <bits/stdc++.h>
using namespace std;
int n, m, k, sum[1005][1005], a[1005][1005];
int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
    {
	for (int j = 1; j <= m; j++)
        {
	    cin >> a[i][j];
	    sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + a[i][j];
	}
    }
    for (int i = 1; i <= k; i++)
    {
        int x1, y1, x2, y2;
	cin >> x1 >> y1 >> x2 >> y2;
	cout << sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1] << endl;
    }
    return 0;
}