lightOJ - 1104（概率与期望）

最新推荐文章于 2021-10-21 11:40:14 发布

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本文探讨了著名的生日悖论，即在一个房间里至少有23人时，两人拥有相同生日的概率超过0.5。通过数学分析，我们进一步讨论了如何根据不同星球的年长度，计算出至少需要邀请多少人，才能使至少两人有相同生日的概率达到0.5。文章提供了详细的算法实现，并通过代码示例进行了解释。

题目：LightOJ - 1104

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

365

669

Sample Output

Case 1: 22

Case 2: 30

题意：在一个房间中，至少有23个人就可以使两个人生日相同的概率大于0.5，现在给出你一年的天数，求出至少需要多少人可以使两个人生日相同的概率至少为0.5；

分析：至少两个人生日相同的概率 = 1 - 所有人生日都不同的概率

所有人生日都不同的概率 P = 1*（n - 1）/n * (n -2)/n * (n - 3)/ n*······*(n - m) / n;

#include <iostream>
#include <stdio.h>

using namespace std;

const int maxn = 100005;


int main()
{
    int T,Case = 0;
    int n;
    cin >> T;
    while(T --)
    {
        cin >> n;
        int cnt = n;
        int ans = 0;
        double p = 1;
        while(p > 0.5)
        {
            cnt --;
            p = p * cnt / n;
            ans ++;
        }
        printf("Case %d: %d\n",++ Case,ans);
    }
    return 0;
}