可能性dp+减少国家HDU4336

Card Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4336

Appoint description:  System Crawler  (2014-10-23)

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 

 

/*************************************************************************
    > File Name: t.cpp
    > Author: acvcla
    > Mail: acvcla@gmail.com 
    > Created Time: 2014年10月21日 星期二 21时33分55秒
 ************************************************************************/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<cstdlib>
#include<ctime>
#include<set>
#include<math.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
double dp[1<<20],p[1<<20],A[200];
void Init(int n){
	memset(p,0,sizeof p);
	memset(dp,0,sizeof dp);
	for(int i=(1<<n)-1;i>0;i--){
		double t=0;
		for(int j=0;j<n;j++)if(1<<j&i)
		{
			p[i]+=A[j];
		}
	}
}
int n;
int main(int argc, char const *argv[])
{
	while(~scanf("%d",&n)){
		double s=1;
		for(int i=0;i<n;i++){
			scanf("%lf",A+i);
			s-=A[i];
		}
		Init(n);
		for(int i=(1<<n)-2;i>=0;i--){
			double t=0;
			double pi=p[i]+s;
			for(int j=0;j<n;j++){
				if((1<<j)&i)continue;
				else{
					int temp=i|(1<<j);
					t+=dp[temp]*A[j];
				}
			}
			dp[i]=(t+1)/(1-pi);
		}
		printf("%.4f\n",dp[0]);
	}
	return 0;
}

版权声明：本文博客原创文章，博客，未经同意，不得转载。