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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
想事先打表、然后输出的，发现空间爆了，坑
思路：每50个单位长度的区间统计一下和，对每次输入的n/50 ，多出的部分就直接一个一个算
#include<cstdio>
#include<cmath>
double num[100000009/50];
int main()
{
num[0]=0;
int ans=1;
for(int i=1; i<=(1e8+50) / 50;++i){
double a=0;
for(int j=(i-1)*50+1;j<=i*50;j++) //每单位长度为50的区间统计一次总和
a+=(1.0/j);
num[i]=num[i-1]+a;
}
int n,t=0;
scanf("%d",&n);
while(n--){
int a;
scanf("%d",&a);
int b=a/50;
double fin=num[b];
for(int i=b*50+1;i<=a;++i) //多出来的的再来单个单个算、
fin+=1.0/i;
printf("Case %d: %.10lf\n",++t,fin);
}
}
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原文地址：http://www.cnblogs.com/sasuke-/p/5239878.html
