hdu 4336 Card Collector (容斥原理)

Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1715    Accepted Submission(s): 786

Special Judge

Problem DescriptionIn 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.

 

InputThe 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.

 

OutputOutput 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

 

Source2012 Multi-University Training Contest 4

題意：

買東西集齊全套卡片贏大獎。每個封裝袋裡面有一張卡片或者沒有。

已知每種卡片出現的機率 p[i]，以及所有的卡片種類的數量 n(1<=n<=20)。

問集齊卡片需要買東西的數量的期望值。

思路：

容斥原理，dfs實現，但是還是不太懂，那個兩個集合的交集應該這麼算: 1/(p[i]+p[j])，以後懂了再解釋。路過的大神也可以給我講一下。

代碼：

#include <iostream>#include <cstdio>#include <cstring>#define maxn 25using namespace std;int n,m;double ans,sum;double a[maxn];void dfs(int len,int pos,int cnt,double s)  // 卡片個數、當前卡片的下標、已經有的卡片、已有卡片的機率{    if(cnt==len)    {        sum+=1.0/s;          return ;    }    if(pos>n) return ;    dfs(len,pos+1,cnt+1,s+a[pos]);    dfs(len,pos+1,cnt,s);}int main(){    int i,j;    while(~scanf("%d",&n))    {        ans=0;        for(i=1; i<=n; i++)        {            scanf("%lf",&a[i]);            ans+=1.0/a[i];        }        for(i=2; i<=n; i++)        {            sum=0;            dfs(i,1,0,0);            if(i%2==0) ans-=sum;  // 減偶            else ans+=sum;        // 加奇        }        printf("%.6f\n",ans);    }    return 0;}