bzoj3132: 上帝造题的七分钟

Posted 2020-07-15

题目链接

嗯，裸的二维区间加区间求和

复习了下zkw的求前缀和的前缀和方法：

维护sx[n] = ∑x[i], six[n] = ∑i * x[i]

那么有ssx[n] = sx[n] * (n + 1) - six[n]

用树状数组搞搞就可以啦

很容易推广到二维

#include <bits/stdc++.h>
using namespace std;
int n, m;
int s1[2100][2100], sn[2100][2100], sm[2100][2100], snm[2100][2100];
int lowbit(int t) {return (t & (-t));}
void add(int s[][2100], int ni, int mi, int a)
{
    for (int i = ni; i <= n; i += lowbit(i))
        for (int j = mi; j <= m; j += lowbit(j))
            s[i][j] += a;
}
int sum(int s[][2100], int ni, int mi)
{
    int a = 0;
    for (int i = ni; i; i -= lowbit(i))
        for (int j = mi; j; j -= lowbit(j))
            a += s[i][j];
    return a;
}
void addx(int ni, int mi, int a)
{
    if (ni * mi == 0) return;
    add(s1, ni, mi, a);
    add(sn, ni, mi, a * ni);
    add(sm, ni, mi, a * mi);
    add(snm, ni, mi, a * ni * mi);
}
int sumx(int ni, int mi)
{
    if (ni * mi == 0) return 0;
    return 
        sum(s1, ni, mi) * (ni + 1) * (mi + 1) - 
        sum(sn, ni, mi) * (mi + 1) -
        sum(sm, ni, mi) * (ni + 1) +
        sum(snm, ni, mi);
}
int main()
{
    char ch[2];
    scanf("%s%d%d", ch, &n, &m);
    while (scanf("%s", ch) == 1)
    {
        int nl, ml, nr, mr, a = 0;
        if (ch[0] == ‘L‘)
        {
            scanf("%d%d%d%d%d", &nl, &ml, &nr, &mr, &a);
            addx(nr + 1, mr + 1, a);
            addx(nl, mr + 1, -a);
            addx(nr + 1, ml, -a);
            addx(nl, ml, a);
        }
        else
        {
            scanf("%d%d%d%d", &nl, &ml, &nr, &mr);
            printf("%d\n", sumx(nr, mr) - sumx(nl - 1, mr) - sumx(nr, ml - 1) + sumx(nl - 1, ml - 1));
        }
    }
}