题解：隔一段数字存一个答案，在查询时，只要找到距离n最近而且小于n的存答案值，再把剩余的暴力跑一遍就可以。
#include <bits/stdc++.h>
using namespace std;
const int N = 1e8 + 10;
const int M = 2e6 + 10;
double a[M];
void Init()
{
a[0] = 0.0;
double ans = 1;
for( int i = 2; i < N; i ++)
{
ans += 1.0 / i;
if(i % 50 == 0)
{
a[i/50] = ans;
}
}
return ;
}
int main()
{
int t,n,cas = 0;
Init();
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
int now = n / 50;
double ans = a[now];
for(int i = now*50 + 1; i <= n; i ++)
{
ans += 1.0 / i;
}
printf("Case %d: %.9lf
",++cas,ans);
}
return 0;
}数论正解：
知识点：
调和级数(即f(n))至今没有一个完全正确的公式，但欧拉给出过一个近似公式：(n很大时)
f(n)≈ln(n)+C+1/2*n
欧拉常数值：C≈0.57721566490153286060651209
c++ math库中，log即为ln。
（转自：https://www.cnblogs.com/shentr/p/5296462.html）
因为公式存在误差，在数值n比较小的时候直接暴力求解。
/** 转自：https://www.cnblogs.com/shentr/p/5296462.html */
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double r=0.57721566490153286060651209; //欧拉常数
double a[10000];
int main()
{
a[1]=1;
for (int i=2;i<10000;i++)
{
a[i]=a[i-1]+1.0/i;
}
int n;
cin>>n;
for (int kase=1;kase<=n;kase++)
{
int n;
cin>>n;
if (n<10000)
{
printf("Case %d: %.10lf
",kase,a[n]);
}
else
{
double a=log(n)+r+1.0/(2*n);
//double a=log(n+1)+r;
printf("Case %d: %.10lf
",kase,a);
}
}
return 0;
}Problem
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