[练习](一二维 )前缀和 与 差分

目录

​​AcWing 795. 前缀和​​

​​AcWing 796.子矩阵的和 ​​

​​AcWing 797.差分​​

​​AcWing 798.差分矩阵​​

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AcWing 795. 前缀和

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输入样例：

5 3
2 1 3 6 4
1 2
1 3
2 4

输出样例：

3
6
10

 

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1e5+10;
int a[N];

int main()
{
    int n,m;
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ ){
        scanf("%d", &a[i]);
        a[i] += a[i-1];
    }
    int l,r;
    for (int i = 0; i < m; i ++ ){
        scanf("%d%d", &l, &r);
        cout << a[r]-a[l-1] << endl;
    }
    
    return 0;
}

AcWing 796.子矩阵的和 

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输入样例：

3 4 3
1 7 2 4
3 6 2 8
2 1 2 3
1 1 2 2
2 1 3 4
1 3 3 4

输出样例：

17
27
21

 

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1010;
int a[N][N],s[N][N];

int main()
{
    int n,m,x;
    cin >> n >> m >> x;
    for (int i = 1; i <= n; i ++ )
        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];
        }
        
    int x1,x2,y1,y2;
    while (x -- )
    {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        cout << s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1] <<endl;
    }
    
    return 0;
}

AcWing 797.差分

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输入样例：

6 3
1 2 2 1 2 1
1 3 1
3 5 1
1 6 1

输出样例：

3 4 5 3 4 2

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1e5+10;
int a[N],b[N];

int main()
{
    int n,m;
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ ){
        scanf("%d", &a[i]);
        b[i]=a[i]-a[i-1];
    }
    int l,r,add;
    for (int i = 0; i < m; i ++ ){
        scanf("%d%d%d", &l, &r, &add);
        b[l] += add;
        b[r+1] -= add;
    }
    
    for (int i = 1; i <= n; i ++ ){
        b[i] += b[i-1];
        printf("%d ",b[i]);
    }
    return 0;
}

AcWing 798.差分矩阵

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输入样例：

3 4 3
1 2 2 1
3 2 2 1
1 1 1 1
1 1 2 2 1
1 3 2 3 2
3 1 3 4 1

输出样例：

2 3 4 1
4 3 4 1
2 2 2 2

 

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1010;
int a[N][N],b[N][N];

void change(int x1,int y1,int x2,int y2,int add)
{
    b[x1][y1] += add;
    b[x1][y2+1] -= add;
    b[x2+1][y1] -= add;
    b[x2+1][y2+1] += add;
}
int main()
{
    int n,m,x;
    cin >> n >> m >>x;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ ){
            scanf("%d", &a[i][j]);
            change(i,j,i,j,a[i][j]);
        }
        
    int x1,x2,y1,y2,add;
    for (int i = 0; i < x; i ++ ){
        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &add);
        change(x1, y1, x2, y2, add);
    }
    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];
            cout << b[i][j] << ' ';
        }
        cout << endl;
    }
    return 0;
}