谐波级数的总和

给定整数N ，任务是找到谐波序列的总和$\sum_{i=1}^{n} \lfloor{n/i}\rfloor$ 。
例子：

Input: N = 5 
Output: 10 
floor(3/1) + floor(3/2) + floor(3/3) = 3 + 1 + 1 = 5
Input: N = 20 
Output: 66 
 

天真的方法：在1到N之间运行一个循环，找到N / i的下限值的总和。这种方法的时间复杂度为O(n) 。
高效方法：使用以下公式来计算序列的总和：
$$\sum_{i=1}^n\lfloor\frac ni\rfloor=2\sum_{i=1}^k\lfloor\frac ni\rfloor-k^2$ for $k=\lfloor\sqrt n\rfloor$$
现在，循环需要从1运行到sqrt(N) ，时间复杂度降低为O(sqrt(N))
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the summation of
// the given harmonic series
long long int getSum(int n)
{
 
    // To store the summation
    long long int sum = 0;
 
    // Floor of sqrt(n)
    int k = sqrt(n);
 
    // Summation of floor(n / i)
    for (int i = 1; i <= k; i++) {
        sum += floor(n / i);
    }
 
    // From the formula
    sum *= 2;
    sum -= pow(k, 2);
 
    return sum;
}
 
// Driver code
int main()
{
    int n = 5;
 
    cout << getSum(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
    // Function to return the summation of
    // the given harmonic series
    static long getSum(int n)
    {
     
        // To store the summation
        long sum = 0;
     
        // Floor of sqrt(n)
        int k = (int)Math.sqrt(n);
     
        // Summation of floor(n / i)
        for (int i = 1; i <= k; i++)
        {
            sum += Math.floor(n / i);
        }
     
        // From the formula
        sum *= 2;
        sum -= Math.pow(k, 2);
     
        return sum;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 5;
     
        System.out.println(getSum(n));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
from math import floor, sqrt, ceil
 
# Function to return the summmation of
# the given harmonic series
def getSum(n):
 
    # To store the summmation
    summ = 0
 
    # Floor of sqrt(n)
    k =(n)**(.5)
 
    # Summation of floor(n / i)
    for i in range(1, floor(k) + 1):
        summ += floor(n / i)
 
    # From the formula
    summ *= 2
    summ -= pow(floor(k), 2)
 
    return summ
 
# Driver code
n = 5
 
print(getSum(n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
     
class GFG
{
     
    // Function to return the summation of
    // the given harmonic series
    static double getSum(int n)
    {
     
        // To store the summation
        double sum = 0;
     
        // Floor of sqrt(n)
        int k = (int)Math.Sqrt(n);
     
        // Summation of floor(n / i)
        for (int i = 1; i <= k; i++)
        {
            sum += Math.Floor((double)n / i);
        }
     
        // From the formula
        sum *= 2;
        sum -= Math.Pow(k, 2);
     
        return sum;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int n = 5;
     
        Console.WriteLine(getSum(n));
    }
}
     
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
 
// Function to return the summation of
// the given harmonic series
function getSum(n)
{
 
    // To store the summation
    let sum = 0;
 
    // Floor of sqrt(n)
    let k = parseInt(Math.sqrt(n));
 
    // Summation of floor(n / i)
    for (let i = 1; i <= k; i++) {
        sum += Math.floor(n / i);
    }
 
    // From the formula
    sum *= 2;
    sum -= Math.pow(k, 2);
 
    return sum;
}
 
// Driver code
    let n = 5;
 
    document.write(getSum(n));
 

输出：

10