LightOJ

题目链接：https://vjudge.net/problem/LightOJ-1104

1104 - Birthday Paradox

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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
2 365 669	Case 1: 22 Case 2: 30

题意：

问至少要多少人，才能使得至少两个人同一天生日的概率不小于0.5？其中一年的天数由题目给定。

题解：

1.至少2个人同一天生日的对立事件为：所有人的生日都不同。先求对立事件，然后再：p(C) = 1-p(!C) 。

2.假设答案为m，如果所有人的生日都不同，那么： p(!C) = A[n][m]/n^m = n/n * (n-1)/n * …… *(n-m+1)/n。

3.m从1开始枚举，当 1- p(!C) >= 0.5 时， m-1就是答案（题目求除己之外）。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e2+100;

int main()
{
    int T, n, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        double p = 1.0;
        int m;
        for(m = 1; ;m++)
        {
            p *= 1.0*(n-m+1)/n;
            if(1-p>=0.5) break;
        }
        printf("Case %d: %d
",++kase, m-1);
    }
}