【算法】前缀和与差分

前缀和

一维前缀和

S[i] = a[1] + a[2] + ... a[i]

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

题目：

解答:

#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;
}

二维前缀和

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]

题目：

解答：

#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;
}

差分

a1,a2 ... an 前缀和

b1,b2 ... bn 差分

使得ai = b1+b2+ ... + bi

一维差分

给区间[l, r]中的每个数加上c：B[l] += c, B[r + 1] -= c

题目：

解答：

#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]);

    while (m -- )
    {
        int l, r, c;
        scanf("%d%d%d", &l, &r, &c);
        insert(l, r, c);
    }

    for (int i = 1; i <= n; i ++ ) b[i] += b[i - 1];

    for (int i = 1; i <= n; i ++ ) printf("%d ", b[i]);

    return 0;
}

二维差分

给以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵中的所有元素加上c：

S[x1, y1] += c, S[x2 + 1, y1] -= c, S[x1, y2 + 1] -= c, S[x2 + 1, y2 + 1] += c

题目：

解答：

#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[x2 + 1][y1] -= c;
    b[x1][y2 + 1] -= 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, y1, x2, y2, c;
        cin >> 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("");  // 换行 等同于printf("\n");
    }

    return 0;
}