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最新推荐文章于 2020-04-16 22:30:13 发布

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本文探讨了一种使用随机选择的方格进行矩阵涂鸦的问题，通过数学计算得出最终涂画的平均数量。针对不同矩阵大小和涂鸦次数，提供了精确的预期涂画数计算方法。

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思路：显然考虑正着做会很麻烦，因为矩阵是可以重叠的，那么就设矩阵k次都没被选中的概率为pp,那么最后累加1-pp就是概率了。对于任意一个矩阵，他不被选的概率为(r-1)^2*m^2+(c-1)^2*n^2+(n-r)^2*m^2+(m-c)^2*n^2，注意到四个角会被重复计算一次，那么再减去(r-1)^2*(c-1)^2,(r-1)^2*(m-c)^2,(n-r)^2*(c-1)^2,(n-r)^2*(m-c)^2就是啦

坑点：500^4会爆int

#include<bits/stdc++.h>
using namespace std;
#define LL long long
int main()
{
    int T;
	scanf("%d",&T);
	int cas = 1;
	while (T--)
	{
		int n,m,k;
		scanf("%d%d%d",&n,&m,&k);
		LL tot = (LL)n*n*m*m;
		double ans = 0;
		for (int r=1;r<=n;r++)
		{
			for (int c = 1;c<=m;c++)
			{
				LL p = 0;
				p += (LL)(r-1)*(r-1)*m*m;
				p += (LL)(c-1)*(c-1)*n*n;
				p += (LL)(n-r)*(n-r)*m*m;
				p += (LL)(m-c)*(m-c)*n*n;
				p -= (LL)(r-1)*(r-1)*(c-1)*(c-1);
				p -= (LL)(n-r)*(n-r)*(c-1)*(c-1);
				p -= (LL)(r-1)*(r-1)*(m-c)*(m-c);
				p -= (LL)(n-r)*(n-r)*(m-c)*(m-c);
				double tempp = (double)p/tot;
				double pp = 1.0;
				for (int i = 1;i<=k;i++)
					pp*=tempp;
				ans+=1.0-pp;
			}
		}
		printf("Case #%d: %.0f\n",cas++,ans);
	}
}

Problem Description

Sakura has a very magical tool to paint walls. One day, kAc asked Sakura to paint a wall that looks like an

M×N matrix. The wall has

M×N squares in all. In the whole problem we denotes

(x,y) to be the square at the

x-th row,

y-th column. Once Sakura has determined two squares

(x1,y1) and

(x2,y2), she can use the magical tool to paint all the squares in the sub-matrix which has the given two squares as corners.

However, Sakura is a very naughty girl, so she just randomly uses the tool for

K times. More specifically, each time for Sakura to use that tool, she just randomly picks two squares from all the

M×N squares, with equal probability. Now, kAc wants to know the expected number of squares that will be painted eventually.

Input

The first line contains an integer

T≤100), denoting the number of test cases.

For each test case, there is only one line, with three integers

M,N and

K.
It is guaranteed that

1≤M,N≤500,

1≤K≤20.

Output

For each test case, output ''Case #t:'' to represent the

t-th case, and then output the expected number of squares that will be painted. Round to integers.

Sample Input

2
3 3 1
4 4 2

Sample Output

Case #1: 4
Case #2: 8

Hint
The precise answer in the first test case is about 3.56790123.