Card Collector
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.
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.
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.
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
Recommend
zhoujiaqi2010
解题思路：有n个卡片，你如今买一包方便面，没包方便面出现当中一个卡片的概率为 p[i] 。问你集齐一套卡片须要的张数的数学期望。
概率DP，用位进制0表示这个卡片有了，1表示这个卡片还没有。那么 比如 “3” 用二进制表示 “1 1” 那么 数组 dp[3] 记录的就是 1号卡片和2号卡片都有的情况集齐一套卡片须要的张数的数学期望。
解题代码：dp[sum]= ( 1+sum { dp[ sum + (1<<j )] *p[j] } ) /sum{p[j] }
当中 ( i&(1<<j) )==0
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=(1<<20)+10;
int n;
double dp[maxn];
double p[30];
void solve(){
int sum=(1<<n)-1;
dp[sum]=0;
for(int i=sum-1;i>=0;i--){
double tmp=0;
dp[i]=1;
for(int j=0;j<n;j++){
if( ( i&(1<<j) )==0 ){
dp[i]+=dp[i+(1<<j)]*p[j];
tmp+=p[j];
}
}
dp[i]/=tmp;
}
printf("%lf\n",dp[0]);
}
int main(){
while(scanf("%d",&n)!=EOF){
for(int i=0;i<n;i++) scanf("%lf",&p[i]);
solve();
}
return 0;
}
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5167129.html，如需转载请自行联系原作者
