hdu 4336 Card Collector （概率dp+位运算 求期望）

题目链接

Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2711    Accepted Submission(s): 1277
Special Judge

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：

有N(1<=N<=20)张卡片，每包中含有这些卡片的概率为p1,p2,````pN.
每包至多一张卡片，可能没有卡片。
求需要买多少包才能拿到所以的N张卡片，求次数的期望。

分析：

n为20，2^20 = 1 048 576;

所以可以用每一位来表示这种卡片有没有存在，还是逆推。

逆推公式：

d[i] = 1.0 + d[i]*p2 + d[i | (1<<j)]*p[j];

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <cmath>
#include <algorithm>
#define LL __int64
const int maxn = (1<<20) + 10;
using namespace std;
double d[maxn], p[30];

int main()
{
    int n, i, j;
    double sum, tmp;
    while(~scanf("%d", &n))
    {
        memset(d, 0, sizeof(d));
        sum = 0;
        for(i = 0; i < n; i++)
        {
            scanf("%lf", &p[i]);
            sum += p[i];
        }
        tmp = 1.0-sum;
        d[(1<<n)-1] = 0;
        for(i = (1<<n)-2; i >= 0; i--)
        {
            double p2 = tmp, p3 = 0;
            for(j = 0; j < n; j++)
            {
                if(i&(1<<j))
                p2 += p[j];  //p2表示没有抽到新的卡片的概率和
                else
                p3 += p[j] * d[i|(1<<j)]; 
            }
            d[i] += (1.0+p3)/(1.0-p2);
        }
        printf("%.4lf
", d[0]);
    }
    return 0;
}