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