HDU 4336：Card Collector（容斥原理）

http://acm.split.hdu.edu.cn/showproblem.php?pid=4336

Card Collector

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：要收集N张卡，吃一包方便面得到第i张卡的概率为p[i]，问收集N张卡吃的方便面包数的期望。

思路：容斥原理。奇数加偶数减。

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

double p[22];
double dp[22];

int main()
{
    int n;
    while(~scanf("%d", &n)) {
        for(int i = 0; i < n; i++)
            scanf("%lf", &p[i]);
        double ans = 0;
        double sum;
        int cnt;
        for(int i = 1; i < (1 << n); i++) {
            sum = 0; cnt = 0;
            for(int j = 0; j < n; j++) {
                if(i & (1 << j)) {
                    cnt++;
                    sum += p[j];
                }
            }
            if(cnt & 1) ans += 1.0 / sum;
            else ans -= 1.0 / sum;
        }
        printf("%f
", ans);
    }
    return 0;
}