一点都不知道这个公式怎么来的？求解释！！！
代码是别人的。答案挺对的。求解公式来的过程,翻译题目就不用。泪！这个题放在高中可能还知道，全忘干净了。
View Code
#include <cstdio> #include <iostream> using namespace std; const int maxn = 22; double p[maxn]; int main() { int n,i,j; while (~scanf("%d",&n)) { for (i = 0; i < n; ++i) scanf("%lf",&p[i]); double ans = 0.0; //按照二项式定理C（n，0）+C（n，1） + ... + C（n，n） = 2^n //所以这里2^n - 1种可能，列举 for (i = 1; i < (1<<n); ++i) { int ct = 0; double tmp = 0.0; for (j = 0; j < n; ++j) { if (i&(1<<j))//搜检0到n中存在于i状况的点 { ct++; tmp += p[j]; cout<<ct<<" "<<tmp<<endl; } } //鸽巢定理 cout<<(ct&1)<<endl; if (ct&1) ans += 1.0/tmp; else ans -= 1.0/tmp; } printf("%.4lf\n",ans); } return 0; }
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 889 Accepted Submission(s): 390
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
