CF2044H Hard Demon Problem

将行和列分开考虑。

在每组询问 $(x_1,y_1,x_2,y_2)$ 中：

• 对于每一行，相邻的两个数的下标差为 $1$。

• 对于每一列，相邻的两个数的下标差为 $y_2-y_1+1$。

不难想到对行和列分别做 $i\times a_{i,j}$ 和 $j\times a_{i,j}$ 的前缀和处理。由于列的下标之差不为 $1$，所以需要先对其做扩大倍数的处理后再将其相加。设最终要求的 $\sum i{a}_i$ 的 $i$ 为系数，那么对于一组询问，可以得到以下系数矩阵：

$$\left[\begin{array}{cccc}{y}_1+x_1(y_2-y_1+1)& y_1+1+x_1(y_2-y_1+1)& \cdots & y_2+x_1(y_2-y_1+1)\\ y_1+(x_1+1)(y_2-y_1+1)& y_1+1+(x_1+1)(y_2-y_1+1)& \cdots & y_2+(x_1+1)(y_2-y1+1)\\ ⋮& ⋮& ⋮& ⋮\\ y_1+x_2(y_2-y_1+1)& y_1+1+x_2(y_2-y_1+1)& \cdots & y_2+x_2(y_2-y_1+1)\end{array}\right]$$

然而正确的系数矩阵应为：

$$\left[\begin{array}{cccc}1& 2& \cdots & y_2-y_1+1\\ y_2-y_1+2& y_2-y_1+3& \cdots & 2(y_2-y_1+1)\\ ⋮& ⋮& ⋮& ⋮\\ (x_2-x_1)(y_2-y_1+1)+1& (x_2-x_1)(y_2-y_1+1)+2& \cdots & (x_2-x_1+1)(y_2-y_1+1)\end{array}\right]$$

将矩阵中的每一个数字都减去 $x_1(y_2-y_1+1)+y_1-1$ 次即可，也就是再维护一个普通的二维前缀和后处理。总时间复杂度为 $O(t(n^2+q))$，代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define init(x) memset (x,0,sizeof (x))
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 2e3 + 5;
const int MOD = 1e9 + 7;
inline int read ();
int t,n,q; 
ll sum[MAX][MAX],sumx[MAX][MAX],sumy[MAX][MAX];
int main ()
{
	//freopen (".in","r",stdin);
	//freopen (".out","w",stdout);
	t = read ();
	while (t--)
	{
		n = read ();q = read ();
		for (int i = 1;i <= n;++i)
		{
			for (int j = 1;j <= n;++j)
			{
				int x = read ();
				sum[i][j] = x + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
				sumx[i][j] = i * x + sumx[i - 1][j] + sumx[i][j - 1] - sumx[i - 1][j - 1];
				sumy[i][j] = j * x + sumy[i - 1][j] + sumy[i][j - 1] - sumy[i - 1][j - 1];;
			}
		} 
		while (q--)
		{
			int sx = read (),sy = read (),fx = read (),fy = read ();
			ll s = sum[fx][fy] - sum[sx - 1][fy] - sum[fx][sy - 1] + sum[sx - 1][sy - 1];
			ll sum_x = sumx[fx][fy] - sumx[sx - 1][fy] - sumx[fx][sy - 1] + sumx[sx - 1][sy - 1];
			ll sum_y = sumy[fx][fy] - sumy[sx - 1][fy] - sumy[fx][sy - 1] + sumy[sx - 1][sy - 1];
			//同一列的公差为 fy - sy + 1
			printf ("%lld ",sum_x * (fy - sy + 1) + sum_y - s * (1ll * sx * (fy - sy + 1) + sy - 1));
		}	
		puts ("");
	}
	return 0;
}
inline int read ()
{
    int s = 0;int f = 1;
    char ch = getchar ();
    while ((ch < '0' || ch > '9') && ch != EOF)
	{
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9')
	{
        s = s * 10 + ch - '0';
        ch = getchar ();
    }
    return s * f;
}