不超过N的所有整数与它们的除数的乘积之和

给定正整数N ，任务是找到范围[1，N]中所有整数与其除数的乘积之和。

例子：

Input: N = 3
Output: 11
Explanation:
Number of positive divisors of 1 is 1( i.e. 1 itself). Therefore, f(1) = 1.
Number of positive divisors of 2 is 2( i.e. 1, 2). Therefore, f(2) = 2.
Number of positive divisors of 3 is 2( i.e. 1, 3). Therefore, f(3) = 2.
So the answer is 1*f(1) + 2*f(2) + 3*f(3) = 1*1 + 2*2 + 3*2 = 11.

Input: N = 4
Output: 23
Explanation:
Here f(1) = 1, f(2) = 2, f(3) = 2 and f(4) = 3. So, the answer is 1*1 + 2*2 + 3*2 + 4*3 = 23.

朴素的方法：朴素的方法是从1遍历到N，并找到所有数字的总和及其除数。

时间复杂度： O(N * sqrt(N))
辅助空间： O(1)

高效方法：要优化上述方法，其想法是考虑总贡献 每个数字都对答案有所帮助。步骤如下：

1. 对于在范围[1，N]的任何数量X，X有助于K，从1之和对于每个k到N，使得K是X的倍数。
2. 观察到这些K的列表的形式为i，2i，3i，…，Fi 其中F是i在1到N之间的倍数。
3. 因此，要找到上面生成的这些数字的总和，则由

i*(1+2+3 + … F) = i*(F*(F+1))/2. 
 

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the sum of the
// product of all the integers and
// their positive divisors up to N
long long sumOfFactors(int N)
{
    long long ans = 0;
 
    // Iterate for every number
    // between 1 and N
    for (int i = 1; i <= N; i++) {
 
        // Find the first multiple
        // of i between 1 and N
        long long frst = i;
 
        // Find the last multiple
        // of i between 1 and N
        long long last = (N / i) * i;
 
        // Find the total count of
        // multiple of in [1, N]
        long long factors
            = (last - frst) / i + 1;
 
        // Compute the contribution of i
        // using the formula
        long long totalContribution
            = (((factors) * (factors + 1)) / 2) * i;
 
        // Add the contribution
        // of i to the answer
        ans += totalContribution;
    }
 
    // Return the result
    return ans;
}
 
// Driver code
int main()
{
    // Given N
    int N = 3;
 
    // function call
    cout << sumOfFactors(N);
}

Java

// Java program for the above approach
class GFG{
   
// Function to find the sum of the
// product of all the integers and
// their positive divisors up to N
static int sumOfFactors(int N)
{
    int ans = 0;
  
    // Iterate for every number
    // between 1 and N
    for (int i = 1; i <= N; i++)
    {
  
        // Find the first multiple
        // of i between 1 and N
        int frst = i;
  
        // Find the last multiple
        // of i between 1 and N
        int last = (N / i) * i;
  
        // Find the total count of
        // multiple of in [1, N]
        int factors = (last - frst) / i + 1;
  
        // Compute the contribution of i
        // using the formula
        int totalContribution = (((factors) *
                                  (factors + 1)) / 2) * i;
  
        // Add the contribution
        // of i to the answer
        ans += totalContribution;
    }
  
    // Return the result
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    // Given N
    int N = 3;
  
    // function call
    System.out.println(sumOfFactors(N));
}
}
 
// This code is contributed by Ritik Bansal

Python3

# Python3 implementation of
# the above approach
 
# Function to find the sum of the
# product of all the integers and
# their positive divisors up to N
def sumOfFactors(N):
 
    ans = 0
 
    # Iterate for every number
    # between 1 and N
    for i in range(1, N + 1):
 
        # Find the first multiple
        # of i between 1 and N
        frst = i
 
        # Find the last multiple
        # of i between 1 and N
        last = (N // i) * i
 
        # Find the total count of
        # multiple of in [1, N]
        factors = (last - frst) // i + 1
 
        # Compute the contribution of i
        # using the formula
        totalContribution = (((factors *
                              (factors + 1)) // 2) * i)
 
        # Add the contribution
        # of i to the answer
        ans += totalContribution
 
    # Return the result
    return ans
 
# Driver Code
 
# Given N
N = 3
 
# Function call
print(sumOfFactors(N))
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
class GFG{
   
// Function to find the sum of the
// product of all the integers and
// their positive divisors up to N
static int sumOfFactors(int N)
{
    int ans = 0;
  
    // Iterate for every number
    // between 1 and N
    for (int i = 1; i <= N; i++)
    {
  
        // Find the first multiple
        // of i between 1 and N
        int frst = i;
  
        // Find the last multiple
        // of i between 1 and N
        int last = (N / i) * i;
  
        // Find the total count of
        // multiple of in [1, N]
        int factors = (last - frst) / i + 1;
  
        // Compute the contribution of i
        // using the formula
        int totalContribution = (((factors) *
                                  (factors + 1)) / 2) * i;
  
        // Add the contribution
        // of i to the answer
        ans += totalContribution;
    }
  
    // Return the result
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    // Given N
    int N = 3;
  
    // function call
    Console.WriteLine(sumOfFactors(N));
}
}
 
// This code is contributed by gauravrajput1

Javascript

// javascript program for the above approach   
// Function to find the sum of the
    // product of all the integers and
    // their positive divisors up to N
    function sumOfFactors(N)
    {
        var ans = 0;
 
        // Iterate for every number
        // between 1 and N
        for (i = 1; i <= N; i++)
        {
 
            // Find the first multiple
            // of i between 1 and N
            var frst = i;
 
            // Find the last multiple
            // of i between 1 and N
            var last = parseInt(N / i) * i;
 
            // Find the total count of
            // multiple of in [1, N]
            var factors = parseInt((last - frst) / i )+ 1;
 
            // Compute the contribution of i
            // using the formula
            var totalContribution = parseInt(((factors) * (factors + 1)) / 2) * i;
 
            // Add the contribution
            // of i to the answer
            ans += totalContribution;
        }
 
        // Return the result
        return ans;
    }
 
    // Driver code
     
        // Given N
        var N = 3;
 
        // function call
        document.write(sumOfFactors(N));
 
// This code is contributed by todaysgaurav.

输出：

11

时间复杂度： O(N)
辅助空间： O(1)