AcWing算法基础课---第一讲基础算法---03前缀和与差分

前缀和

思路：求l到r区间的和用前r个数减去前l - 1个数.

#include <iostream>

using namespace std;

const int N = 100010;
int a[N], s[N];

int main()
{
    int n, m;
    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];
    
    while (m --)
    {
        int l, r;
        scanf("%d%d", &l, &r);
        printf("%d\n", s[r] - s[l - 1]);
    }
    
    return 0;
}

Acwing 796. 子矩阵的和

#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]; //求前缀和
            
    while (q --)
    {
        int x1, x2, y1, 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]); //求部分和
    }
    
    return 0;
}

AcWing 797. 差分

#include <iostream>

using namespace std;

const int N = 100010;
int n, m;
int a[N], b[N];

void insert(int l, int r, int c)
{
    b[l] += c;
    b[r + 1] -= c;
}

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 ++) insert(i, i, a[i]); //求差分数组b[N]
    
    while (m --)
    {
        int l, r, c;
        scanf("%d%d%d", &l, &r, &c);
        insert(l, r, c); //差分数组插入处理
    }
    
    for (int i = 1; i <= n; i ++) b[i] += b[i - 1]; //求前缀和还原数组
    
    for (int i = 1; i <= n; i ++) printf("%d ", b[i]);
    
    return 0;
    
}

AcWing 798. 差分矩阵

#include <iostream>

using namespace std;

const int N = 1010;

int n, m, q;
int a[N][N], b[N][N];

void insert(int x1, int y1, int x2, int y2, int c)
{
    b[x1][y1] += c;
    b[x1][y2 + 1] -= c;
    b[x2 + 1][y1] -= c;
    b[x2 + 1][y2 + 1] += c;
    
}

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 ++)
            insert(i, j, i, j, a[i][j]);
            
    while (q --)
    {
        int x1, x2, y1, y2, c;
        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &c);
        insert(x1, y1, x2, y2, c);
    }
    
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];
            
    
    for (int i = 1; i <= n; i ++)
    {
        for (int j = 1; j <= m; j ++) printf("%d ", b[i][j]);
        puts(" ");
    }
        
            
    return 0;   

}