Acwing算法学习之前缀和

Posted 2020-12-14 qwer0553

前缀和十分好理解，与数学上面的等差数列类同

先上前缀和模板

以下是一维前缀和

公式如下

S[i] = a[1] + a[2] + ... a[i]
a[l] + ... + a[r] = S[r] - S[l - 1]

二维前缀和模板

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]

模板题

Acwing 795.前缀和
来源 : https://www.acwing.com/problem/content/797/
直接套模板，没有什么难点

#include <iostream>

using namespace std;

const int N = 100001 + 10;
int q[N], s[N], n, m;

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ ) cin >> q[i], s[i] = s[i - 1] + q[i];
    
    
    while (m -- )
    {
        int l, r;
        cin >> l >> r;
        
        cout << s[r] - s[l - 1] << endl;
    }
    
    return 0;
}

二维前缀和模板题

Acwing 796.子矩阵的和
来源 ： https://www.acwing.com/problem/content/798/

#include <iostream>

using namespace std;

const int N = 1000 + 10;
int a[N][N], s[N][N], n, m, k;

int main()
{
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ ) 
            cin >> 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]; // 通过图像来记忆
    while (k -- )
    {
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        
        cout << s[x2][y2] - s[x2][y1 - 1] - s[x1 - 1][y2] + s[x1 - 1][y1 - 1] << endl;
    }
    
    return 0;
}