概率dp HDU 4336 Card CollectorCard Collector

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概述

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

Sample Output


10.000
10.500

Source

2012 Multi-University Training Contest 4

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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。

思路：卡片种类最多20种，第一时间想到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，我们用dp[s] 表示从s到集齐的期望步数，那么dp[s] = sigma(dp[t]*p[j])+sigma(dp[s]*p[j])+1; 然后我们只要分开状态不变的系数和状态转移的系数，那么就可以直接算出dp[s]

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define eps 1e-9
const int maxn = 1<<21;
double p[22];
double f[maxn];
int n;
void init()
{
memset(f,0,sizeof(f));
}
void input()
{
for (int i = 0 ; i < n ; ++i) scanf("%lf",p+i);
}
void solve()
{
double pp;
for (int s = (1<<n)-2 ; s >= 0 ; --s)
{
pp = 0;
for (int j = 0 ; j < n ; ++j)
{
if (s&(1<<j)) continue;
int t = s+(1<<j);
f[s] += p[j]*f[t];
pp += p[j];
}
f[s] = (f[s]+1)/pp;
}
printf("%lfn",f[0]);
}
int main()
{
while (scanf("%d",&n)==1)
{
init();
input();
solve();
}
}