hdu4336 Card Collector

最新推荐文章于 2021-06-24 22:54:17 发布

转载最新推荐文章于 2021-06-24 22:54:17 发布 · 63 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/code-painter/p/3751054.html

本文介绍了一个经典的概率问题——通过购买零食收集卡片直至集齐所有种类的数学期望计算方法。使用容斥原理解决此问题，并提供了完整的C++代码实现。

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

0.1

0.1 0.4

Sample Output

10.000

10.500

题目大意：给出N个奖品，每买一个物品得到第i个奖品的概率为ai，求收集齐所有的奖品需要买多少物品的数学期望。

思路：简单第容斥原来。用i表示取奖品第状态，按个数判断加或减

 1 #include<cstdio>
 2 
 3 int main()
 4 {
 5     int n;
 6     double a[21];
 7     while (~scanf("%d",&n))
 8     {
 9         for (int i=0;i<n;++i)
10             scanf("%lf",&a[i]);
11         double temp,ans=0;
12         for (int i=1;i< (1<<n);++i)
13         {
14             temp=0;
15             int ct=0;
16             for (int j=0;j<n;++j)
17               if (i & (1<<(j) ) )
18               {
19                   ct++;
20                   temp+=a[j];
21               }
22             if (ct & 1) ans+=1/temp;
23             else ans-=1/temp;
24         }
25         printf("%.4f\n",ans);
26     }    
27     return 0;
28 }

转载于:https://www.cnblogs.com/code-painter/p/3751054.html