light oj 1284 - Lights inside 3D Grid(求期望)
2015-07-23 16:21
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1284 - Lights inside 3D Grid
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.
Each case starts with a line containing four integers X, Y, Z (1 ≤ X, Y, Z ≤ 100) and K (0 ≤ K ≤ 10000).
在一个三维的网格中 三维长度各为X Y Z
在网格中有一些灯 一开始这些灯都是关着的
进行k次操作 每次操作随机选择两个点 以这两个点为边界的三维图形中的灯状态取反
求经过k次操作之后 亮着的灯的期望值
对于某个点(i,j,k) 先求处该点被选择的概率 对于每一维 只要选择的两个点在(i,j,k)的同一侧就不会被选到
在一维中(以x维为例) 选择两个点总的可能数是x*x 不会被选到的可能数是(i-1)*(i-1)+(x-i)*(x-i)
两者相减 得到被选择的可能数
假设f(n)为进行n次操作 被选择奇数次的期望 g(n)为被选择偶数次的期望
f(n)=(1-p)*f(n-1)+p*g(n-1)
可以推导出一个等比数列
PDF (English) | Statistics | Forum |
Time Limit: 4 second(s) | Memory Limit: 32 MB |
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 |
在网格中有一些灯 一开始这些灯都是关着的
进行k次操作 每次操作随机选择两个点 以这两个点为边界的三维图形中的灯状态取反
求经过k次操作之后 亮着的灯的期望值
对于某个点(i,j,k) 先求处该点被选择的概率 对于每一维 只要选择的两个点在(i,j,k)的同一侧就不会被选到
在一维中(以x维为例) 选择两个点总的可能数是x*x 不会被选到的可能数是(i-1)*(i-1)+(x-i)*(x-i)
两者相减 得到被选择的可能数
假设f(n)为进行n次操作 被选择奇数次的期望 g(n)为被选择偶数次的期望
f(n)=(1-p)*f(n-1)+p*g(n-1)
可以推导出一个等比数列
#include <cstdio> #include <iostream> #include <cstring> #include <cmath> #include <algorithm> #include <string.h> #include <string> #include <vector> #include <queue> #define MEM(a,x) memset(a,x,sizeof a) #define eps 1e-8 #define MOD 10009 #define MAXN 10010 #define MAXM 100010 #define INF 99999999 #define ll #define bug cout<<"here"<<endl #define fread freopen("ceshi.txt","r",stdin) #define fwrite freopen("out.txt","w",stdout) using namespace std; int Read() { char c = getchar(); while (c < '0' || c > '9') c = getchar(); int x = 0; while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); } return x; } void Print(int a) { if(a>9) Print(a/10); putchar(a%10+'0'); } double num(int a,int b) { double all=1.0*b*b; double x=1.0*(a-1)*(a-1)+1.0*(b-a)*(b-a); return all-x; } int main() { //fread; int tc; int cs=1; scanf("%d",&tc); while(tc--) { int x,y,z,K; scanf("%d%d%d%d",&x,&y,&z,&K); double ans=0.0,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 k=1;k<=z;k++) { double p=num(i,x)*num(j,y)*num(k,z)/all; ans+=0.5-0.5*pow(1.0-2.0*p,1.0*K); } printf("Case %d: %.7lf\n",cs++,ans); } return 0; }
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