Problem:
Score : 400 points

Problem Statement

For a positive integer X, let f(X) be the number of positive divisors of X.

Given a positive integer N, find \sum_{K=1}^N K\times f(K).

Constraints

1 \leq N \leq 10^7

Input

Input is given from Standard Input in the following format:

N

Output

Print the value \sum_{K=1}^N K\times f(K).

Sample Input 1

4

Sample Output 1

23

We have f(1)=1, f(2)=2, f(3)=2, and f(4)=3, so the answer is 1\times 1 + 2\times 2 + 3\times 2 + 4\times 3 =23.

Sample Input 2

100

Sample Output 2

26879

Sample Input 3

10000000

Sample Output 3

838627288460105

Watch out for overflows.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
// using namespace std;
int N;
int i = 2;
long long num;
char buf[8], out[20];
char k;
s;
size_t ii;
int main()
{
  s = read(0, buf, 8);
  for (;;)

  {
    k = buf[ii++];
    if (k < '0')
      break;
    N = N * 10 + k - '0';
  }
  // printf("%d\n", N);
  long long ans = (long long)N * (N + 1) - 1;

  for (; i * i <= N; i++)
  {
    num = N / i;
    // ans += (long long)(num - i + 1) * i * (i + num) - i * i;
    ans += (num * num - i * i + num) * i;
  }
  printf("%lld\n", ans);
}