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 <math.h>
#include <stdio.h>
int n, sqrtN;
long long ans, d;
char buf[8],out[20];
char k;
s;
size_t i;
int main() {
    s = read(0, buf, 8);
    for(;;) {
        k = buf[i++];
        if(k < '0') break;
        n = n * 10 + k - '0';
    }
    if(n == 1) {
        putchar(49);
    } else {
        const int sqrtN = sqrt(n);
        ans = -(long long)(sqrtN) * (sqrtN + 1) * (2 * sqrtN + 1) / 6;
        for(int i = 1; i <= sqrtN; ++i) {
            const int a = n / i;
            ans += ((long long)i * (i + a) * (a + 1 - i));
        }
        i = 20;
        do{
            out[--i]=ans%10+'0';
            ans/=10;
        }while(ans);
        write(1,out+i,20-i);
    }
}