hdu 4336 Card Collector （概率与期望+状压DP）

Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4305    Accepted Submission(s): 2168
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

10.120.1 0.4

 

Sample Output

10.00010.500

 

Source

2012 Multi-University Training Contest 4

 

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题解：概率与期望+状压DP

还是需要逆着推，f[(1<<n)-1]=0

然后枚举j,判断这个位置是否已经出现过来

f[i]=(f[i|(1<<j)]+1)*p[j]+(f[i]+1)*p[k] j表示的是i状态中没出现的，k表示的是i状态中出现过的以及空包的概率。

然后移项，求f[i]即可。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int n;double f[1<<21],p[23];int main(){freopen("a.in","r",stdin);while (scanf("%d",&n)!=EOF) {memset(f,0,sizeof(f)); double p1=1.0;for (int i=0;i<n;i++) scanf("%lf",&p[i]),p1-=p[i];//cout<<p1<<endl;f[(1<<n)-1]=0;for (int i=(1<<n)-2;i>=0;i--) {f[i]=0;int t=i; double sum=p1;for (int j=0;j<n;j++) if ((t>>j)&1) sum+=p[j]; else  f[i]+=(f[i|(1<<j)]+1)*p[j];    f[i]+=sum; //cout<<f[i]<<" "<<(1.0-sum)<<endl;    f[i]/=(double)(1.0-sum);}printf("%.4lf\n",f[0]);}}