Card Collector(HDU 4336)

Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3001    Accepted Submission(s): 1435
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

 

Source

2012 Multi-University Training Contest 4

 

Recommend

zhoujiaqi2010

求期望

方法一：状压

逆序枚举所有状态 d[i] 表示状态为i时收集完所有卡片的期望步数。

d[i] = 1 + ∑(d[i | (1 << j)] * p[j])(ps: 累加所有走一步会增加新一张卡片的期望步数) + (1 - t) * d[i](ps: t为增加一张新卡片的概率);

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define MAXN (1 << 20)

double d[MAXN + ];
double p[];
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
       memset(d, , sizeof(d));
       repu(i, , n) scanf("%lf", &p[i]);
       if(( << n) - ) d[( << n) - ] = 0.0;
       double t;
       for(int i = ( << n) - ; i >= ; i--)
       {
           d[i] += 1.0;
           t = 0.0;
           for(int j = ; j < n; j++)
              if(!(i & ( << j))) {
                d[i] += p[j] * d[i | ( << j)];
                t += p[j];
              }
           d[i] /= t;
       }
       printf("%.4lf\n", d[]);
    }

    return ;
}

方法二：容斥

设Ai表示取到第i张卡片的期望，Ai = 1 / pi;

由容斥原理得：

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define MAXN (1<<20)

double p[];
double d[MAXN + ];
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        double re = 0.0;
        repu(i, , n) scanf("%lf", &p[i]);
        int m = ;
        double t = 0.0;
        repu(i, , ( << n)) {
           m = , t = 0.0;
           repu(j, , n) if(i & ( << j)) t += p[j], m++;
           if(m & ) re += 1.0 / t;
           else      re -= 1.0 / t;
        }
        printf("%.4lf\n", re);
    }
    return ;
}