Birthday Paradox LightOJ - 1104

标签：common   out   problems   remember   ems   1.0   double   style   font   

Sometimes some mathematical results are hard to believe. One of the common problems is the birthday paradox. Suppose you are in a party where there are 23 people including you. What is the probability that at least two people in the party have same birthday? Surprisingly the result is more than 0.5. Now here you have to do the opposite. You have given the number of days in a year. Remember that you can be in a different planet, for example, in Mars, a year is 669 days long. You have to find the minimum number of people you have to invite in a party such that the probability of at least two people in the party have same birthday is at least 0.5.

Input

Input starts with an integer T (≤ 20000), denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤ 10⁵) in a single line, denoting the number of days in a year in the planet.

Output

For each case, print the case number and the desired result.

Sample Input

2

365

669

Sample Output

Case 1: 22

Case 2: 30

题解：好像暴力都能过。。。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn=1e5+5;
 5 
 6 int kase,n,ans;
 7 double dp[maxn];
 8 
 9 int main()
10 {   cin>>kase;
11     for(int t=1;t<=kase;t++){
12         cin>>n;     //用scanf("%d",&n)能少900多毫秒。。。。
13         double temp=1.0;
14         for(int i=1;i<=n;i++){
15             temp*=(n-i)*1.0/n;
16             if(temp<=0.5){ ans=i; break; } 
17         }
18         printf("Case %d: %d\n",t,ans);
19     }
20     return 0;
21 } 

Birthday Paradox LightOJ - 1104

标签：common   out   problems   remember   ems   1.0   double   style   font   

原文地址：http://www.cnblogs.com/zgglj-com/p/7787510.html