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 <stdlib.h>
#include <math.h>
#include <string.h>

//gcdやり直す

void swap (int *x, int *y) {
  int temp;    

  temp = *x;
  *x = *y;
  *y = temp;
}




/***
* pivotを決め、
* 全データをpivotを境目に振り分け、
* pivotの添え字を返す
***/
int partition (int array[], int left, int right) {
  int i, j, pivot;
  i = left;
  j = right + 1;
  pivot = left;   // 先頭要素をpivotとする

  do {
    do { i++; } while (array[i] < array[pivot]);
    do { j--; } while (array[pivot] < array[j]);
    // pivotより小さいものを左へ、大きいものを右へ
    if (i < j) { swap(&array[i], &array[j]); }
  } while (i < j);

  swap(&array[pivot], &array[j]);   //pivotを更新

  return j;
}

/* クイックソート */
void quick_sort (int array[], int left, int right) {
  int pivot;

  if (left < right) {
    pivot = partition(array, left, right);
    quick_sort(array, left, pivot-1);   // pivotを境に再帰的にクイックソート
    quick_sort(array, pivot+1, right);
  }
}




/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/

int main(void){
  long int n;
  scanf("%ld",&n);
  long int ans=0;
  for(int i=1;i<=n;i++){
    long int now=n/i;
    ans+=now*(now+1)/2*i;
  }

  printf("%ld\n",ans);
  return 0;
}