题链：
题意
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
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()); } }}