前缀和作用：

快速求出原数组中一段数组的和 

思路

1.预处理前缀和数组

2.用公式求区间和

公式：

二维前缀和：

---

s [ i ] [ j ] += s[ i - 1 ] [ j ] + s[ i ] [ j - 1 ] - s [ i - 1 ] [ j - 1]; 

 

 题型

一维

二维

 题解

一维

#include <iostream>using namespace std;const int N = 100010;int n, m;
int a[N], s[N];int main()
{scanf("%d%d", &n, &m);for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);for (int i = 1; i <= n; i ++ ) s[i] = s[i - 1] + a[i]; // 前缀和的初始化while (m -- ){int l, r;scanf("%d%d", &l, &r);printf("%d\n", s[r] - s[l - 1]); // 区间和的计算}return 0;
}

二维

#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("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);}return 0;
}