Light OJ 1234 Harmonic Number
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 number 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
思路:没思路啊 这个是高数的东西 发散 n足够大时它无穷大 直接公式解
#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <algorithm> #include <iostream> using namespace std; const int maxn = 1000010; const double phi = 0.57721566490153286060651209;///这个数字就是后来称作的欧拉常数。不过遗憾的是,我们对这个常量还知之甚少,连这个数是有理数还是无理数都还是个谜。但是此题中所用是ln(n+0.5)+phi;double a[maxn];
int main() { for(int i = 1; i <= 1000000; i++) a[i] = a[i-1] + 1.0/i; int cas = 1; int T; scanf("%d", &T); while(T--) { int n; scanf("%d", &n); double ans; if(n <= 1000000) ans = a[n]; else ans = log(n+0.5) + phi; printf("Case %d: %.10lf\n", cas++, ans); } return 0; }