LightOJ - 1284 Lights inside 3D Grid —— 期望

题目链接：https://vjudge.net/problem/LightOJ-1284

 

1284 - Lights inside 3D Grid

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Time Limit: 4 second(s)

Memory Limit: 32 MB

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^-6 will be ignored.

Sample Input	Output for Sample Input
5 1 2 3 5 1 1 1 1 1 2 3 0 2 3 4 1 2 3 4 2	Case 1: 2.9998713992 Case 2: 1 Case 3: 0 Case 4: 6.375 Case 5: 9.09765625

 

 

 

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e5+100;

double Get(int pos, int n)
{
    return 1.0 - 1.0*((pos-1)*(pos-1)+(n-pos)*(n-pos))/(n*n);
}

double qpow(double x, int y)
{
    double s = 1;
    while(y)
    {
        if(y&1) s *= x;
        x *= x;
        y >>= 1;
    }
    return s;
}

int main()
{
    int T, x, y, z, k, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        double 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 t = 1; t<=z; t++)
                {
                    double p = Get(i,x)*Get(j,y)*Get(t,z);
                    ans += 0.5-0.5*qpow(1-2*p, k);
                }
        printf("Case %d: %.10lf\n", ++kase, ans);
    }
}