hdoj4336 Card Collector

题目：

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
1
0.1
2
0.1 0.4
Sample Output
10.000
10.500

题意就是给你需要收集的卡片以及收集每种卡片的概率，问需要买多少个包才能收集齐（期望）。
分类：概率dp

其实我也不太明白概率dp的原理， 先把代码打上。

参考代码：

#include 
#include 
#include 
#include 
using namespace std;
const int MAXN = 20+2;
//0: 这种卡片没有收集到; 1: 这种卡片收集到了;
double p[MAXN];
double dp[1<= 0;--i) {//从还有一件没有集齐开始计算;
            double x = 0, sum = 1;
            for (int j = 0;j < n;++j) {//枚举每一种卡片;
                if (i & (1<