概率dp HDU4336 Card Collector

最新推荐文章于 2019-09-11 15:23:28 发布

丘大路

最新推荐文章于 2019-09-11 15:23:28 发布

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分类专栏：动态规划

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本文探讨了一个关于收集特定数量卡片的问题，通过购买零食包获得卡片，每个包中可能包含不同类型的卡片之一，或者完全不含卡片。文章提供了一段C++代码实现，用于计算收集完整套卡片所需的预期零食包数量。

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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

这题说实话我也没咋看懂，只是把我的一些想法贴注释里吧，不要当真，也许全想错了（话说来个大牛告诉sum到底是什么意思啊啊啊啊）

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
using namespace std;

const int MAXN=21;
double p[MAXN];
double dp[1<<MAXN];

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i,j;
		double t=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf",&p[i]);
			t+=p[i];
		}
		t=1-t;									//t就表示没有卡片的概率了
		dp[(1<<n)-1]=0;							//全部收集到了就不需要再买了.求期望一般都是反着推.
		for(i=(1<<n)-2;i>=0;i--)
		{
			double sum=1,pp=0;
			for(j=0;j<n;j++)
			{
				if(i&(1<<j))
					pp+=p[j];
				else
					sum+=p[j]*dp[i|(1<<j)];		//sum表示期望值,加上去的表示通过那种取法还差多少才能达到dp[i|(1<<j)]的要求
			}
			dp[i]=sum/(1-t-pp);					//dp[i]*p是期望值
		}
		printf("%.5lf\n",dp[0]);
	}
	return 0;
}