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1234 - Harmonic Number

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Time Limit: 3 second(s)

Memory Limit: 32 MB

In mathematics, the nth harmonic number isthe 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 nthharmonic number. Errors less than
10-8 will be ignored.

Sample Input	Output for Sample Input
12 1 2 3 4 5 6 7 8 9 90000000 99999999 100000000	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

题意：不说了。

分析：一亿的数据量，可以每50个或者100个记录下来。

#include<bitset>
#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

inline int in()
{
int res=0;char c;
while((c=getchar())<'0' || c>'9');
while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
return res;
}
const int N=1e8;
double a[N/50+50];
void init()
{
double t=1.0;
for(int i=2;i<=N;i++)
{
t += 1.0/i;
if(i%50 == 0)
{
a[i/50]=t;
}
}
}
int main()
{
int T=in(),ii=1;
init();
while(T--)
{
int n=in();
double ans = a[n/50];
for(int i=n/50*50+1;i<=n;i++)
{
ans += 1.0/i;
}
printf("Case %d: %.10lf\n",ii++,ans);
}
return 0;
}