hdu 4336 容斥原理 2012 Multi-University Training Contest 4

Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4592    Accepted Submission(s): 2319
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

10.120.1 0.4

 

Sample Output

10.00010.500

 

Source

2012 Multi-University Training Contest 4

题意：

给出n个卡片出现的概率，卡片都是单独出现，不能一次出现多张

题解：

对于第一个样例，出现第一张的期望为 1.0/.01   出现第二张的期望为 1/0.4 

此时我们就想用容斥原理，就要减去在一包中可能出现这两张的概率（不是同时出现，只是可能出现这两张的概率）是 0.1 + 0.4

然后减去这个期望值  1/ (0.1 + 0.4)

很容易发现这就是一个容斥了

然后稍微搞一搞就过了

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define MAXN 100double a[MAXN],ans;int n;void dfs(int now,double num,int sym){    ans+=1.0/num*sym;    for(int i=now+1;i<=n;i++)        dfs(i,num+a[i],-1*sym);}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%lf",&a[i]);        ans=0;        for(int i=1;i<=n;i++)            dfs(i,a[i],1);        printf("%0.4lf\n",ans);    }    return 0;}