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最新推荐文章于 2022-07-30 23:34:34 发布

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最新推荐文章于 2022-07-30 23:34:34 发布

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分类专栏：思维数学

本文链接：https://blog.youkuaiyun.com/qq_34681949/article/details/52204848

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本文介绍了一个基于生日悖论的问题，即在一个包含特定数量的人群中，至少两人拥有相同生日的概率达到0.5所需的最少人数。文章提供了一段C语言代码，用于计算在不同天数的年份中这一临界值的具体数值。

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I - I

Time Limit: 2000MS Memory Limit: 32768KB 64bit IO Format: %lld & %llu

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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 b

irthday? 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

思路：

两个人生日相同的情况太多，用逆向思维，通过循环先算出任意两人生日不同的概率，若概率值小于等于0.5就可以了，结束循环；

计算任意两人生日不同的概率，用一个for循环，先算第一个人的是n*1.0 / n,第二个人的是(n-1) *1.0 / n,第三个人的是（n-2) *1.0 / n 等等；i从0开始，总人数为i+1(包括他自己），要邀请的人数是 i ;

代码：

#include<stdio.h>
#include<string.h>
int main()
{
	int t,n,mm=1;
	scanf("%d",&t);
	while(t--)
	{
		double ans=1.0;
		scanf("%d",&n);
		int i;
		for(i=0;;i++)
		{
			ans*=(n-i)*1.0/n;
			if(ans<=0.5)
			break;
		}
		printf("Case %d: %d\n",mm++,i);
	}
	return 0;
}