hdu 4336 Card Collector[期望]

最新推荐文章于 2021-08-11 10:35:58 发布

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最新推荐文章于 2021-08-11 10:35:58 发布

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分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/code_or_code/article/details/44617397

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本文探讨了儿童时期收集零食卡牌的行为，并通过数学模型计算了收集完整套卡牌所需的平均购买次数。

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Card Collector

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

思路：状态压缩，从后往前推；

f [ s ]=1+f [ s ]*(1-segma[p. i ] ) + f[s]*segma[p. j ] + f[s|1<<k ]*segma[p.k] ;

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#define INF 0x7fffffff
#define SUP 0x80000000
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

typedef long long LL;
const int N=100007;

double a[30],dp[5000000];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        double tt=0,sum;
        for(int i=0;i<n;i++)
        {
            scanf("%lf",a+i);
            tt+=a[i];
        }
        tt=1-tt;   //空包的概率
        dp[(1<<n)-1]=0;
        for(int i=(1<<n)-1-1;i>=0;i--)
        {
            dp[i]=1;
            sum=0;
            for(int j=0;j<n;j++)
            {
                if(i&1<<j) sum+=a[j];
                else dp[i]+=dp[i|1<<j]*a[j];
            }
            dp[i]/=1-tt-sum;
        }
        printf("%.4f\n",dp[0]);
    }
    return 0;
}