参看资料:
https://blog.csdn.net/wl16wzl/article/details/82858903
题目:
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
题目大意:
一个长宽高为 x,y,z 的立方网格之中,每一个节点上都有一个开关,初始状态都为关;现在有 k 次选择,每次随机选择两点,将两点确定的立方体之间 的 所有的点都按一次开关,k 次之后,求灯开着的数目的期望。
解题思路:
单独计算每个点被选中奇数次的期望,加起来就是总亮灯数的期望。
对于一次选取,x轴位置 i 的点,被选中的概率就是1.0减去不被选中的概率【即两个点选中的点都在 i 一侧】
P(x,i)=1.0-1.0*((i-1)*(i-1)+(x-i)*(x-i))/(x*x) ;
三维坐标下被选中的概率就是三个维度概率乘起来p=P(x,i)*P(y,j)*P(z,k);
假设选取n次,f(n)为选取n次被选中奇数次的概率,g(n)为选区n次被选中偶数次的概率
f(n)=p*g(n-1)+(1-p)*f(n-1)
g(n)=p*f(n-1)+(1-p)*g(n-1)
f(n)+g(n)=1
得f(n)=(1-2p)*f(n-1)+p,f(1)=p
展开后为f(n)=(1-2p)^(n-1)*p+(1-2p)^(n-2)*p+......+(1-2p)*p+p
就是一个等比数列,第一项为p,用等比求和公式得f(n)=0.5-0.5*(1-2p)^n
f(n)就是这个点灯亮的期望
实现代码:
#include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;
double P(int i,int x){
return 1.0-1.0*((i-1)*(i-1)+(x-i)*(x-i))/(x*x);
}
double pow1(double a,int b){
double ret=1;
while(b){
if(b&1)ret=ret*a;
a*=a;
b>>=1;
}
return ret;
}
int main(){
int T,x,y,z,k;
scanf("%d",&T);
for(int cas=1;cas<=T;cas++){
scanf("%d%d%d%d",&x,&y,&z,&k);
double ans=0;
for(int i=1;i<=x;i++){
for(int j=1;j<=y;j++){
for(int s=1;s<=z;s++){
double p=P(i,x)*P(j,y)*P(s,z);
ans+=0.5-0.5*pow1(1-2*p,k);
}
}
}
printf ("Case %d: %.7f\n",cas,ans);
}
return 0;
}