N 数的除数之和

Question

Find then sum all the divisors from 1 to N.
The main problem is, that this code runs really poor with high numbers.

The following code was taken from: https://www.geeksforgeeks.org/sum-divisors-1-n

        static int divisorSum(int n)
        {
            int sum = 0;

            for (int i = 1; i <= n; ++i)
            {
                for (int j = 1; j * j <= i; ++j)
                {
                    if (i % j == 0)
                    {
                        if (i / j == j)
                            sum += j;
                        else
                            sum += j + i / j;
                    }
                }
            }
            return sum;
        }

Answer 1

Based on @Joel solution, I just improved it:

static long divisorSum(int n)
{
   long sum = 0;
   for (long i = 1; i <= n/2; ++i)
      sum += i * (n / i);
   sum += (n/2+1+n)*(n-n/2)/2; // It's a sum of an arithmetic progression‏‏‏‏
   return sum;
}

For i > n/2 the expression i * (n / i) is simply i (because n/i = 1), so we can get the sum of all the numbers between n/2 + 1 to n by computing a sum of an arithmetic progression‏‏‏‏. It will run faster, although it is O(n) too.

Answer 2

There is no need to use a collection of sorts, since you sum everything up and don't need to think about duplicates. I don't think there is a way to get a solution for this which is a perfect O(n) , but this is the closest I can think of:

int sum = 0;

for (int i = 1; i <= n; i++) 
{
    double sqrt = Math.Sqrt (i);

    for (int j = 1; j <= sqrt; j++) 
    {
        if (i % j == 0) 
        {
            sum += j;

            if (j != sqrt)
                sum += i / j;
        }
    }
}

Divisors are pairs, so there isn't a need to go all the way to i every time (eg 1 * 10 , 10 * 1 are the same). You can go till the square-root of i (the 'mid-point'), and save time, hence it's not O(n^2) , but not perfectly O(n) .

Answer 3

You could do something like this O(n) :

static long divisorSum(int n)
{
    long sum = 0;
    for (long i = 1; i <= n; ++i)
        sum += i * (n / i);
    return sum;
}

static void Main(string[] args)
{
    int val = 129999;
    Console.WriteLine(divisorSum(val));
    Console.ReadLine();
}

Tests:

12999 => 8ms
129999 => 25ms
2147483647 => 18770ms (Max Int32 value)

---

int val = 129999;
int maxInt = int.MaxValue;

//val (129999)
var watch = System.Diagnostics.Stopwatch.StartNew();
Console.WriteLine(divisorSum(val));
watch.Stop();
var elapsedMs = watch.ElapsedMilliseconds;
Console.WriteLine(elapsedMs); //25ms

//MaxInt (2147483647)
watch = System.Diagnostics.Stopwatch.StartNew();
Console.WriteLine(divisorSum(maxInt));
watch.Stop();
elapsedMs = watch.ElapsedMilliseconds; //18770ms 
Console.WriteLine(elapsedMs);
Console.ReadLine();