Card Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2195    Accepted Submission(s): 1034 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   |   We have carefully selected several similar problems for you:  4337 4331 4332 4333 4334  题目大意：要集齐N张卡片，每包干脆面出现每种卡片的概率已知，问你集齐N张卡片所需要的方便面包数的数学期望 思路：DP[mask]表示已经收集到的卡片距离收集完所有卡片的天数的数学期望 DP[mask[ = DP[mask]*P1+SUM（DP[mask|1<<j]*P）+1 #include <iostream> #include <cstdio> #include <cstring> #include <vector> using namespace std; const int maxn = 1<<20; double dp[maxn]; double p[30]; int n; int main(){ while(cin >> n){ for(int i = 0; i < n; i++) scanf("%lf",&p[i]); int d = (1<<n)-1; memset(dp,0,sizeof dp); dp[d] = 0; for(int i = d-1; i >= 0; i--){ double t = 1.0,tp = 0.0; for(int j = 0; j < n ; j++){ if((i&(1<<j))==0){ t += dp[i|(1<<j)]*p[j]; tp += p[j]; } } dp[i] = t/tp; } printf("%.4lf\n",dp[0]); } return 0; }