【UVA 11181】（条件概率）

题链：https://cn.vjudge.net/problem/UVA-11181

题意

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n个人去了超市，已知每个人买东西的概率为p[i]，在已知有r个人买了东西的情况下，求实际上每个人买东西的概率

题解

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设r个人买东西的时间为E

\[ans=p(i\;|\;E)=\frac{p(iE)}{p(E)} \]

每个人买东西的概率是独立的，在一种r情况下，利用乘法原理即可。
多种r情况是互斥的，累加起来即可。

参考代码

import java.io.*;
import java.util.*;

public class Main {
	static final int N=(int)5005;
	static double p[]=new double[25];
	static boolean vis[]=new boolean[25];
	static int n,r;
	static double sum[]=new double[25];
	static void dfs(int step,int cnt,double res) {
		if(step==n+2||cnt>r) return;
		if(step==n+1&&cnt==r) {
			sum[0]+=res;
			for(int i=1;i<=n;i++) if(vis[i]) {
				sum[i]+=res;
			}
			return;
		}
		vis[step]=false;
		dfs(step+1,cnt,res*(1-p[step]));
		vis[step]=true;
		dfs(step+1,cnt+1,res*p[step]);
	}
    public static void main(String[] args) {  
        InputStream sys=System.in;
        InputReader in=new InputReader(sys);
      //  Scanner sc=new Scanner(new InputStreamReader(sys));
        PrintWriter out=new PrintWriter(System.out);
        int T=1;
        while(true) {
        	n=in.nextInt();r=in.nextInt();
        	if(n==0&&r==0) break;
        	for(int i=1;i<=n;i++) {
        	     p[i]=in.nextDouble();
        	     vis[i]=false;sum[i]=0;
        	}
        	sum[0]=0;
        	dfs(1,0,1);
        	StringBuffer ans=new StringBuffer();
        	for(int i=1;i<=n;i++) {
        	      ans.append(String.format("%.6f\n", sum[i]/sum[0]));	
        	}
        	out.println("Case "+(T++)+":");
        	out.print(ans);
        	out.flush();
        }
    }
	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;
        
		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public double nextDouble() {
			return Double.parseDouble(next());
		}
	}
}