LightOJ 1104 Birthday Paradox（生日悖论，思维）InputOutputSample InputOutput for Sample Input

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概述

http://lightoj.com/volume_showproblem.php?problem=1104

1104 - Birthday Paradox

PDF (English)

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

365

669

Case 1: 22

Case 2: 30

题意：

在一个有 n 天的星球上，邀请一些人来参加聚会，

两个人生日相同的概率 >0.5 ，问至少需要请多少人。

参考思路：百度百科

所以所有人生日都不相同的概率是：

那么，n个人中有至少两个人生日相同的概率就是：

所以当n=23的时候，概率为0.507

当n=100的时候，概率为0.999999692751072

Code：

#include<cstdio
#include<cstring>
const int MYDD = 1103;
int main() {
int TT;
scanf("%d",&TT);
int Kcase=1;
while(TT--) {
int n,ans=1;
scanf("%d",&n);
double p=1;
for(int j=n-1; j>=0; j--) {
p=p*(j*1.0)/(n*1.0);
if(p<=0.5) {
break;
}
ans++;
//
printf("**** p: %lfn",p);
}
printf("Case %d: %dn",Kcase++,ans);//漏掉了 Kcase++
//
printf("%lf",364.0*363.0/(365.0*365.0));
}
return 0;
}
/*
By: Shyazhut */