算生日相同的概率

Description

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 is669 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

    分析：

         为了防止溢出，我们就一边乘一边除！算出生日不在同一天的概率，然后再用一减去就好了

          ps:50 人的party生日相同概率竟然高达93%！！！！！！

#include<iostream>
using namespace std;
int n;
int birthday(int x)
{
    double c = 1.0;
    int m = 0;
    for (int i = 0;; i++)
    {
        c *= (double)(n - i) / n;    //防止溢出
        m++;
        if (1.0 - c >= 0.5)
            break;
    }
    return m;
}
int main()
{
    int T, ans = 1;
    cin >> T;
    while (T--)
    {

        cin >> n;
        cout << "Case " << ans++ << ": " << birthday(n) - 1 << endl;      //要减去本人
    }
    return 0;
}