概率dp HDU 4336 Card Collector

Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1980    Accepted Submission(s): 925
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

 

Recommend

zhoujiaqi2010   |   We have carefully selected several similar problems for you:  4337 4331 4332 4333 4334 

题意：你想要收起N个不同的卡片，给出每一次获得某卡片的概率，然后要求要集齐所有卡片的期望次数。

分析：卡片种类最多20种，可以用状态压缩来表示卡片的获得情况，1代表已经有了，0代表还没有，dp[s]代表集齐s种卡片的期望，s是二进制张开后代表的卡片种类，s从1带2^n-1,有转移：dp[s]=sigma(dp[t]*p[j])+(1-sigma(p[j]))*dp[s]+1  ====>dp[s]=(sigma(dp[t]*p[j])+1)/sigma(pj]);

#include <iostream>#include <cmath>#include <stdio.h>#include <map>#include <string>#include <string.h>#include <algorithm>#include <queue>using namespace std;#define REP(i,a,b)              for(int i=a;i<b;++i)#define scan(a)                 scanf("%d",&a)#define maxn                    (1<<21)#define mset(a,b)               memset(a,b,sizeof a)#define LL                      long long//#define mod                     10000000000000000int N;double dp[maxn];double p[23];int main(){    //dp[0]=0;    while(~scan(N))    {        mset(dp,0);        REP(i,0,N)        {            scanf("%lf",p+i);        }        int nn=(1<<N);        REP(i,1,nn)        {            int s=i;            int t=0;            double pp=0;            for(;s;s/=2,t++)            {                if(s%2 == 0) {continue;}                int j=(1<<t);                int ss=i-j;                dp[i]+=p[t]*dp[ss];                pp+=p[t];            }            dp[i]=(dp[i]+1)/pp;        }        printf("%lf\n",dp[nn-1]);    }}