标签：概率dp

Problem G
Probability|Given
Input: Standard Input

Output: Standard Output

N friends go to the local super market together. The probability of their buying something from the market is respectively. After their marketing is finished you are given the information that exactly r of them has bought something and others have bought nothing. Given this information you will have to find their individual buying probability.

Input

The 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  (0.1<<1) which actually denotes the buying probability of the i-th friend. All probability values should have at most two digits after the decimal point. 

Input is terminated by a case where the value of N and r is zero. This case should not be processes.  

Output

For 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 values should have six digits after the decimal point. Follow the exact format shown in output for sample input. Small precision errors will be allowed. For reasonable precision level use double precision floating-point numbers.

Sample Input                             Output for Sample Input

3 2 

0.10 

0.20 

0.30 

5 1 

0.10 

0.10 

0.10 

0.10 

0.10 

0 0

Problem-setter: Shahriar Manzoor
Special Thanks: Derek Kisman

题意：有n个人，并且知道每个人买东西的概率，现在已知有r个人买了东西，依次求第i个人买东西的概率

思路：ans[i]保存有r个人买了东西的情况下并且第i个人买了东西的概率，all代表n给人中第r个买了东西的概率

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
//typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 25

double ans[N];
double p[N];
int n,r;

double dfs(int pos,int left,double pp)
{
    if(pos>n) return left==0 ? pp : 0;

    double sum=0;

    if(left)
	{
	    sum+=dfs(pos+1,left-1,pp*p[pos]);
	    ans[pos]+=sum;
	}
    sum+=dfs(pos+1,left,pp*(1-p[pos]));
    return sum;
}

int main()
{
	int i,j,ca=0;

    while(sff(n,r),n+r)
	{
		fre(i,1,n+1)
		  scanf("%lf",&p[i]);

		memset(ans,0,sizeof(ans));

	   double all=dfs(1,r,1);
	   pf("Case %d:\n",++ca);
	   for(i=1;i<=n;i++)
			pf("%.6lf\n",ans[i]/all);
	}

  return 0;
}

Uva 11181 Probability|Given（概率dp）

标签：概率dp

原文地址：http://blog.csdn.net/u014737310/article/details/45318815