AcWing 796. 子矩阵的和

题目描述

输入一个 n 行 m 列的整数矩阵，再输入 q 个询问，每个询问包含四个整数 x1,y1,x2,y2，表示一个子矩阵的左上角坐标和右下角坐标。

对于每个询问输出子矩阵中所有数的和。

输入格式

第一行包含三个整数 n，m，q。

接下来 n 行，每行包含 m 个整数，表示整数矩阵。

接下来 q 行，每行包含四个整数 x1,y1,x2,y2，表示一组询问。

输出格式

共 q 行，每行输出一个询问的结果。

数据范围

1≤n,m≤1000,

1≤q≤200000,

1≤x1≤x2≤n1,

1≤y1≤y2≤m1,

−1000≤矩阵内元素的值≤1000

输入样例：

3 4 3
1 7 2 4
3 6 2 8
2 1 2 3
1 1 2 2
2 1 3 4
1 3 3 4

输出样例：

17
27
21

思路

C

#include <iostream>
using namespace std;
int main() {
    int n, m, q;
    scanf("%d%d%d", &n, &m, &q);
    int s[n   1][m   1];
    for (int i = 1; i <= n; i  ) {
        for (int j = 1; j <= m; j  ) {
            int tmp;
            scanf("%d", &tmp);
            s[i][j] = s[i][j - 1]   s[i - 1][j] - s[i - 1][j - 1]   tmp;
        }
    }
    while (q--) {
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        printf("%dn", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1]   s[x1 - 1][y1 - 1]);
    }
}

#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int 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", &s[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];
    while (q--) {
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        printf("%dn", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1]   s[x1 - 1][y1 - 1]);
    }
    return 0;
}

Go

package main
import (
    "bufio"
    "fmt"
    "os"
)
func main() {
    reader := bufio.NewReader(os.Stdin)
    var n, m, q int
    fmt.Fscan(reader, &n, &m, &q)
    s := make([][]int, n 1)
    for i := range s {
        s[i] = make([]int, m 1)
    }
    for i := 1; i <= n; i   {
        for j := 1; j <= m; j   {
            var tmp int
            fmt.Fscan(reader, &tmp)
            s[i][j] = s[i-1][j]   s[i][j-1] - s[i-1][j-1]   tmp
        }
    }
    writer := bufio.NewWriter(os.Stdout)
    defer writer.Flush()
    for i := 0; i < q; i   {
        var x1, y1, x2, y2 int
        fmt.Fscan(reader, &x1, &y1, &x2, &y2)
        result := s[x2][y2] - s[x1-1][y2] - s[x2][y1-1]   s[x1-1][y1-1]
        fmt.Fprintln(writer, result)
    }
}

模板

S[i, j] = 第i行j列格子左上部分所有元素的和
以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵的和为：
S[x2, y2] - S[x1 - 1, y2] - S[x2, y1 - 1]   S[x1 - 1, y1 - 1]