Card Collector

http://acm.hdu.edu.cn/showproblem.php?pid=4336

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Special Judge

Problem Description

In your childhood, do you crazy for collecting the
beautiful cards in the snacks? They said that, for example, if you collect all
the 108 people in the famous novel Water Margin, you will win an amazing award.

As a smart boy, you notice that to win the award, you must buy much more
snacks than it seems to be. To convince your friends not to waste money any
more, you should find the expected number of snacks one should buy to collect a
full suit of cards.

Input

The first line of each test case contains one integer N
(1 <= N <= 20), indicating the number of different cards you need the
collect. The second line contains N numbers p1, p2, ..., pN, (p1 + p2 + ... + pN
<= 1), indicating the possibility of each card to appear in a bag of snacks.

Note there is at most one card in a bag of snacks. And it is possible
that there is nothing in the bag.

Output

Output one number for each test case, indicating the
expected number of bags to buy to collect all the N different cards.

You
will get accepted if the difference between your answer and the standard answer
is no more that 10^-4.

Sample Input

1

0.1

2

0.1 0.4

Sample Output

10.000

10.500

Source

2012
Multi-University Training Contest 4

题意：

n类卡片，买一包零食可能得到一张卡片，可能没有，问最后集齐n类卡片期望买几包零食

题解：

每一包零食，有3种可能

1、没有卡片

2、有 没有收集过的卡片

3、有 已经收集了的卡片

将卡片拥有情况状态压缩成s，p[j]表示没有收集过的卡片的概率

dp[s]=1+ 没有卡的概率*dp[s]+ 收集过的卡片的概率和*dp[s] +  Σ dp[(1<<j)|s]*p[j]

=1+ dp[s]*(没有卡的概率+收集过的卡片的概率和) +  Σ dp[(1<<j)|s]*p[j]

=1+ dp[s]*（1-Σ p[j]）+  Σ dp[(1<<j)|s]*p[j]

移项，dp[s]-dp[s]*（1-Σ p[j]）= 1+ Σ dp[(1<<j)|s]*p[j]

∴dp[s]* Σ p[j]=1+ Σ dp[(1<<j)|s]*p[j]

∴ dp[s]= （1+ Σ dp[(1<<j)|s]*p[j]）/ Σ p[j]

#include<cstdio>
#include<cstring>
using namespace std;
int n;
double p[20],dp[1<<20];
double sum1,sum2;
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++) scanf("%lf",&p[i]);
        for(int i=(1<<n)-2;i>=0;i--)
        {
            sum1=sum2=0;
            for(int j=0;j<n;j++)
             if(!((1<<j)&i))
             {
                 sum1+=p[j];
                 sum2+=dp[(1<<j)|i]*p[j];
             }
            dp[i]=(sum2+1)/sum1;
        }
        printf("%.4lf\n",dp[0]);
    }
}