【前缀和】

一维前缀和

s[l~r]=s[r]-s[l-1]

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=100010;
int a[N],s[N];
int n,m;

int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++) {
        cin>>a[i];
        s[i]=s[i-1]+a[i];
    }
    while(m--){
        int l,r;
        cin>>l>>r;
        printf("%d\n",s[r]-s[l-1]);
    }
    return 0;
}

二维前缀和

计算前缀和

s[x][y]=s[x-1][y]+s[x][y-1]-s[x-1][y-1]+a[x][y]

计算子矩阵和

s[x₁,y₁][x₂,y₂]=s[x₂,y₂]-s[x₂,y₁-1]-s[ x₁-1,y₂]+s[x₁-1,y₁-1]

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010;

int n, m, q;
int a[N][N], s[N][N];

int main()
{
    scanf("%d%d%d", &n, &m, &q);

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
        {
            scanf("%d", &a[i][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;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
    }

    return 0;
}