hdu4336Card Collector

题意：有n种卡片，可以通过买干脆面收集卡片，每包干脆面最多一张卡片，问收集完n种卡片时买的干脆面包数的期望。

做法：

把当前手中有的卡片种类状压成s，

然后dp[s]，状态为s时的期望。

考虑每次买一包干脆面，有三种情况：

1、已经拥有

2、没有拥有

3、没有卡片

于是dp[s]=dp[s]*(1跟3的概率之和)+dp[s|(1<<i)]*p[i],i为还没有的卡片编号，p[i]为摸到它的概率

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
double dp[1<<20],p[20];
int main()
{
	int n;
	while(cin>>n)
	{
		double sum=1;
		for(int i=0;i<n;i++)
		{
			cin>>p[i];
			sum-=p[i];
		}
		dp[(1<<n)-1]=0;
		for(int i=(1<<n)-2;i>-1;i--)
		{
			double s1=0,s2=0;
			for(int j=0;j<n;j++)
				if((i>>j)&1)
					s1+=p[j];
				else
					s2+=dp[i|(1<<j)]*p[j];
			dp[i]=(1+s2)/(1-(s1+sum));
		}
		printf("%.4f\n",dp[0]);
	}
}

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

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