Light Oj 1284 Lights inside 3D Grid（概率期望）

题意：X*Y*Z。每个位置(x,y,z)有一盏灯，初始时是灭的。K轮操作。每轮任意选两盏灯(可以相同)将其之间的所有灯翻转。问最后明着的灯的数量期望。

思路：先求出每盏灯在一轮中被改变的概率，对于(i,j,k)位置的，只要选择的两盏灯不在同一侧就行。那么可以对于每一维分别算，比如对于X这一维，就是减去在其同侧的。然后计算K轮被改变奇数次的概率就是该盏灯最后对答案的贡献。这f(n)表示n次改变奇数次，g(n)表示偶数次，则有f(n)=f(n-1)*(1-p)+g(n-1)*p,然后g(n)=1-f(n)得到：f(n)=f(n-1)*(1-2p)+p，最后得到f(n)=(1-(1-2p)^n)/2。

#include<bits/stdc++.h>
using namespace std;
double cal(int x,int y)
{
	return 1.0*y*y-(x-1)*(x-1)-(y-x)*(y-x);
}
int main()
{
	int T,cas=1;
	scanf("%d",&T);
	while(T--)
	{
         int x,y,z,k;
		 scanf("%d%d%d%d",&x,&y,&z,&k);
         double ans = 0.0;
		 double all = 1.0*x*x*y*y*z*z;
		 for (int i = 1;i<=x;i++)
		 {
			 for (int j = 1;j<=y;j++)
			 {
				 for (int kk = 1;kk<=z;kk++)
				 {
					 double p = 1.0*cal(i,x)*cal(j,y)*cal(kk,z)/all;
					 ans+=(0.5-0.5*pow(1-2*p,1.0*k));
				 }
			 }
		 }
		 printf("Case %d: %lf\n",cas++,ans);
	}
}

Description

You are given a 3D grid, which has dimensions X, Y and Z. Each of the X x Y x Z cells contains a light. Initially all lights are off. You will have K turns. In each of the K turns,

1.      You select a cell A randomly from the grid,

2.      You select a cell B randomly from the grid and

3.      Toggle the states of all the bulbs bounded by cell A and cell B, i.e. make all the ON lights OFF and make all the OFF lights ON which are bounded by A and B. To be clear, consider cell A is (x₁, y₁, z₁) and cell B is (x₂, y₂, z₂). Then you have to toggle all the bulbs in grid cell (x, y, z) where min(x₁, x₂) ≤ x ≤ max(x₁, x₂),min(y₁, y₂) ≤ y ≤ max(y₁, y₂) and min(z₁, z₂) ≤ z ≤ max(z₁, z₂).

Your task is to find the expected number of lights to be ON after K turns.

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with a line containing four integers X, Y, Z (1 ≤ X, Y, Z ≤ 100) and K (0 ≤ K ≤ 10000).

Output

For each case, print the case number and the expected number of lights that are ON after K turns. Errors less than10^-6 will be ignored.

Sample Input

5

1 2 3 5

1 1 1 1

1 2 3 0

2 3 4 1

2 3 4 2

Sample Output

Case 1: 2.9998713992

Case 2: 1

Case 3: 0

Case 4: 6.375

Case 5: 9.09765625