AcWing 796. 子矩阵的和
原题链接
简单
作者:
现代修士o_O
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2021-04-21 08:46:37
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所有人可见
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阅读 267
#include <iostream>
using namespace std;
const int N = 1010;
int n, m, q;
int a[N][N], s[N][N];//初始化前缀和数组,s[0][0..N - 1] = 0, s[0..N - 1][0] = 0;
int main()
{
cin >> n >> m >> q;
for (int i = 1; i <= n; i ++ )//从下标1开始,优雅省事
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]; // 初始化前缀和数组
while(q -- )
{
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
printf("%d\n",s[x2][y2] - s[x2][y1 - 1] - s[x1 - 1][y2] + s[x1 - 1][y1 - 1]); // 求区间部分和
}
return 0;
}