数组an a1 a2 ... an
前缀和 Si = a1 + a2 + ... + ai
①如何求
②作用
// 一维数组前缀和的计算
#include <iostream>
using namespace std;
const int N = 100010;
int a[N], s[N];
int n, m;
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;
}
// 二位数组矩阵前缀和计算
#include <iostream>
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]);
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] + 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;
}
标签:const,前缀,求解,int,scanf,d%,c++,数组
From: https://blog.51cto.com/u_16492348/9284794