LightOJ1284 Lights inside 3D Grid (概率DP)

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 than 10^-6will 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

题解：

参考代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int T,X,Y,Z,K;
double ans;

double calc(int x,int m)
{
    return 1.0-1.0*((m-x)*(m-x)+(x-1)*(x-1))/(m*m);
}

int main()
{
    scanf("%d",&T);
    for(int cas=1;cas<=T;++cas)
    {
        ans=0;
        scanf("%d%d%d%d",&X,&Y,&Z,&K);
        for(int i=1;i<=X;++i)
            for(int j=1;j<=Y;++j)
                for(int k=1;k<=Z;++k)
                {
                    double p=calc(i,X)*calc(j,Y)*calc(k,Z);
                    ans+=0.5-0.5*pow(1.0-2*p,K);
                }
        printf("Case %d: %.8lf\n",cas,ans);
    } 

    return 0;
}

原文地址：https://www.cnblogs.com/songorz/p/10828658.html