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 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
2
365
669
Sample Output
Case 1: 22
Case 2: 30
竞赛时别人都写出来了，我竟没写出来，想复杂了， 拿到一个题要从战略上重视它，从战术上藐视它。 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> using namespace std; int main() { int T,p=0; scanf("%d",&T); while(T--) { int n,i; scanf("%d",&n); double cnt=1.0; for(i=0; i<n; i++) { cnt=cnt*(1.0*n-1.0*i)/n*1.0; //不知道cnt放后面乘为什么结果不对 if(cnt<=0.5) break; } printf("Case %d: %d\n",++p,i); } return 0; }
