HDU - 4336 Card Collector

最新推荐文章于 2021-06-24 22:54:17 发布

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最新推荐文章于 2021-06-24 22:54:17 发布

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分类专栏：训练文章标签： dp

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本文探讨了通过购买零食收集特定卡片的问题，采用状态压缩动态规划方法计算收集全套装所需的预期次数。

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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

N个物品，每次得到第i个物品的概率为pi，而且有可能什么也得不到，问期望多少次能收集到全部N个物品。（有中文题解真好）

求期望标准倒推，i由i+1转化来，dp[sum]= ( 1+sum { dp[ sum + (1<<j )] *p[j] } ) /sum{p[j] }，即i的期望可以为所有此位为0的以往转化来。

这题用了状态压缩dp，状压do大致是当单个状态很简单（如只有是否两种），总数又不多时可以使用二进制来保存每个状态。

AC代码：

#include <bits/stdc++.h>

using namespace std;

const int maxn=(1<<20)+10;

int n;
double p[25];
double dp[maxn];
void solve()
{
    int sum=(1<<n)-1;
    dp[sum]=0;
    for(int i=sum-1;i>=0;i--)
    {
        double tmp=0;
        dp[i]=1;
        for(int j=0;j<n;j++)
        {
            if((i&(1<<j))==0)
            {
                dp[i]+=dp[i+(1<<j)]*p[j];
                tmp+=p[j];
            }
        }
        dp[i]/=tmp;
    }
    printf("%lf\n",dp[0]);
}
int main()
{
    while(cin>>n)
    {
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
            cin>>p[i];
        solve();
    }
    return 0;
}