http://acm.hust.edu.cn:8080/judge/contest/view.action?cid=8086#problem/D
Description
In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:
In this problem, you are given n, you have to find Hn.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 108).
Output
For each case, print the case nu
mber and the nth harmonic number. Errors less than 10-8 will be ignored.
Sample Input
12
1
2
3
4
5
6
7
8
9
90000000
99999999
100000000
Sample Output
Case 1: 1
Case 2: 1.5
Case 3: 1.8333333333
Case 4: 2.0833333333
Case 5: 2.2833333333
Case 6: 2.450
Case 7: 2.5928571429
Case 8: 2.7178571429
Case 9: 2.8289682540
Case 10: 18.8925358988
Case 11: 18.9978964039
Case 12: 18.9978964139
#include <iostream> #include <stdio.h> #include <math.h> using namespace std; #define M 0.57721566490153286060651209 double b[10000]; int main() { int n,i; b[1]=1; for (i=2;i<10000;i++) { b[i]=b[i-1]+1.0/i; } cin>>n; for (i=1;i<=n;i++) { int m; cin>>m; if (m<10000) { printf("Case %d: %.10lf\n",i,b[m]); } else { double a=log(m)+M+1.0/(2*m); printf("Case %d: %.10lf\n",i,a); } } return 0; }