Probability|Given - UVa 11181 概率dp

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提比-我有特殊的AC技巧

最新推荐文章于 2022-06-17 15:56:27 发布

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分类专栏： UVa dp动态规划和递推

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本文介绍了一道关于概率计算的问题，给出了AC代码实现过程。利用动态规划求解给定条件下朋友购买商品的概率。

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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-thfriend. 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

Case 1:

0.413043

0.739130

0.847826

Case 2:

0.200000

题意：n个小伙伴中有m个买了东西，告诉你每次小伙伴单独买东西的概率，问每个小伙伴买了东西的概率。

思路：设Ai为第i个小伙伴买了东西，B为小伙伴们恰好买了m个东西，P(A|B)=P(AB)/P(B)。通过dp来求前n个小伙伴中有m个小伙伴买东西的概率。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
double dp[2][25],p[25];
int t,n,m;
int main()
{
    int i,j,k,a,b,f;
    double ret;
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0 && m==0)
          break;
        for(i=1;i<=n;i++)
           scanf("%lf",&p[i]);
        printf("Case %d:\n",++t);
        if(m==0)
        {
            for(i=1;i<=n;i++)
               printf("0.000000\n");
            continue;
        }
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        a=0;b=1;
        for(i=1;i<=n;i++)
        {
            dp[b][0]=dp[a][0]*(1-p[i]);
            for(j=1;j<=i;j++)
               dp[b][j]=dp[a][j-1]*p[i]+dp[a][j]*(1-p[i]);
            a^=1;b^=1;
        }
        ret=dp[a][m];
        for(k=1;k<=n;k++)
        {
            memset(dp,0,sizeof(dp));
            dp[0][0]=1;
            a=0;b=1;
            for(i=1;i<=n;i++)
               if(i!=k)
               {
                   dp[b][0]=dp[a][0]*(1-p[i]);
                   for(j=1;j<=i;j++)
                      dp[b][j]=dp[a][j-1]*p[i]+dp[a][j]*(1-p[i]);
                   a^=1;b^=1;
               }
            printf("%.6f\n",dp[a][m-1]*p[k]/ret);
        }

    }
}