# HDU 4336 Card Collector （容斥原理||概率DP）

### 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 thecollect. 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

### 思路

#### 概率dp

• 每一个状态都是由比它小的其他状态通过改变一位二进制数转移得到的
• 从 dp[0] 一直计算到 dp[2^n-1] 。
• 最后一个全 1 状态的值即获得所有卡片的期望数量。

### 容斥原理

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double a[25];
void solve(int N)
{
double ans=0.0;
for(int msk=1; msk<(1<<N); msk++)
{
double sum=0;
int flag=0;
for(int i=0; i<N; i++)
{
if(msk&(1<<i))  //选中一个二进制位
{
++flag;
sum+=a[i];
}
}
if(flag&1)ans+=1.0/sum; //奇加偶减
else ans-=1.0/sum;
}
printf("%lf\n",ans);
}

int main()
{
int N;
while(~scanf("%d",&N))
{
for(int i=0; i<N; i++)
scanf("%lf",a+i);
solve(N);
}
return 0;
}

### 概率dp

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include<iostream>
using namespace std;
double a[25],dp[1<<25];
void solve(int T)
{
int maxx=(1<<T)-1;
dp[0]=0;
for(int i=1; i<=maxx; i++)
{
double tmp=0;
dp[i]=1;    //当前状态一定会成功1*P/P
for(int j=0; j<T; j++)
{
if((1<<j)&i)    //找到1位
{
dp[i]+=dp[i-(1<<j)]*a[j];   //计算前项期望*转移概率
tmp+=a[j];  //计算概率和
}
}
dp[i]/=tmp;
}
printf("%lf\n",dp[maxx]);
}
int main()
{
int T;
while(~scanf("%d",&T))
{
for(int i=0; i<T; i++)
scanf("%lf",a+i);
solve(T);
}
return 0;
}