F - 概率（经典问题）

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 are23 people including you. What is the probability that at least two people in the party have same birthday? Surprisingly the result is more than0.5. 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 least0.5.

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

题意：

假设一年有n天，你要找你的朋友来参加party，且要求至少有两个人生日相同的概率大于等于0.5的最少人数，这个就是你要邀请的最小人数。

思路：

求至少有两个人生日相同的情况实在有很多，所以我们逆向思维，求出任意两个人生日都不相同的概率，当达到要求是就退出循环，算的的结果就是最小人数，要注意的是

我们这里要减去自己，应为是要邀请的人数，还要注意防止数据溢出，因此需要边乘边除。参照算法竞赛与入门经典P324-325就能解决这道题目。

代码：

#include<cstdio>double birthday(int n){    double ans=1.0;    int m=0;    for(int i=0;; i++)    {        ans*=(double)(n-i)/n;        m++;        if(1.0-ans>=0.5)            break;    }    return m;}int main(){    int T,n,m;    int ans;    scanf("%d",&T);    int test=1;    while(T--)    {        scanf("%d",&n);        ans=birthday(n)-1;        printf("Case %d: %d\n",test++,ans);    }    return 0;}