全概率......
Description 
Problem G Output: Standard Output
N friends go to the local super market together. The probability of their buying something from the market is
InputThe input file contains at most 50 sets of inputs. The description of each set is given below:
First line of each set contains two integers N (1 ≤ N ≤ 20) and r(0 ≤ r ≤ N). Meaning of N and r are given in the problem statement. Each of the next N lines contains one floating-point number
Input is terminated by a case where the value of N and r is zero. This case should not be processes.
OutputFor each line of input produce N+1 lines of output. First line contains the serial of output. Each of the next N lines contains a floating-point number which denotes the buying probability of the i-th friend given that exactly r has bought something. These
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respectively. After their marketing is finished you
(0.1<
3 2 0.10 0.20 0.30 5 1 0.10 0.10 0.10 0.10 0.10 0 0 |
Case 1: 0.413043 0.739130 0.847826 Case 2: 0.200000 0.200000 0.200000 0.200000 0.200000
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Problem-setter: Shahriar Manzoor
Special Thanks: Derek Kisman
Source
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 6. Mathematical Concepts and Methods
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 2. Mathematics :: Probability :: Exercises:
Beginner
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Probability
Theory
Root :: Prominent Problemsetters :: Shahriar Manzoor
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 5. Mathematics :: Probability
Theory
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Probability Theory :: Standard
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int n,r;
double p[50];
double ans[50];
int cot(int x)
{
int ret=0;
while(x)
{
x=x&(x-1);
ret++;
}
return ret;
}
int main()
{
int cas=1;
while(scanf("%d%d",&n,&r)!=EOF)
{
if(n==0&&r==0) break;
for(int i=0;i<n;i++)
scanf("%lf",p+i);
memset(ans,0,sizeof(ans));
double all=0.0;
for(int i=0;i<(1<<n);i++)
{
if(cot(i)==r)
{
double ppp=1.;
for(int j=0;j<n;j++)
{
if(i&(1<<j)) ppp*=p[j];
else ppp*=(1-p[j]);
}
all+=ppp;
for(int j=0;j<n;j++)
{
if(i&(1<<j)) ans[j]+=ppp;
}
}
}
printf("Case %d:\n",cas++);
for(int i=0;i<n;i++)
{
printf("%.6lf\n",ans[i]/all);
}
}
return 0;
}