LightOJ - 1234 LightOJ - 1245 Harmonic Number（欧拉系数+调和级数）

Harmonic Number

In mathematics, the n^th 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 H_n.

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 n^th harmonic number. Errors less than 10^-8 will be ignored.

Sample Input

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

求1+1/2+1/3+...+1/n

#include<bits/stdc++.h>

#define e 0.57721566490153286060651209

using namespace std;

typedef long long ll;

double a[];

int main()

{

    int tt=,t,n,i;

    a[]=;

    for(i=;i<=;i++){

        a[i]=a[i-]+1.0/i;

    }

    scanf("%d",&t);

    while(t--){

        scanf("%d",&n);

        if(n<=){

            printf("Case %d: %.10lf\n",++tt,a[n]);

        }

        else{

            double ans=log(n)+e+1.0/(*n);

            printf("Case %d: %.10lf\n",++tt,ans);

        }

    }

    return ;

}

LightOJ - 1234

Harmonic Number (II)

I was trying to solve problem '1234 - Harmonic Number', I wrote the following code

long long H( int n ) {
  long long res = 0;
  for( int i = 1; i <= n; i++ )
res = res + n / i;
  return res;
}

Yes, my error was that I was using the integer divisions only. However, you are given n, you have to find H(n) as in my code.

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n < 2³¹).

Output

For each case, print the case number and H(n) calculated by the code.

Sample Input

2147483647

Sample Output

Case 1: 1

Case 2: 3

Case 3: 5

Case 4: 8

Case 5: 10

Case 6: 14

Case 7: 16

Case 8: 20

Case 9: 23

Case 10: 27

Case 11: 46475828386

求n+n/2+n/3+...+n/n

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int main()

{

    int tt=,T,i;

    ll n;

    cin>>T;

    while(T--)

    {

        cin>>n;

        int m=sqrt(n);

        ll sum=;

        for(i=;i<=m;i++){

            sum+=n/i;

        }

        for(i=;i<=m;i++){

            sum+=(n/i-n/(i+))*i;

        }

        if(m==n/m) sum-=m;

        cout<<"Case "<<++tt<<": "<<sum<<endl;

    }

    return ;

}