[Nowcoder 218396] Little G's sum

Small G's sum

Title link: nowcoder 218396

To Niu Ke:

——>Click me to jump<——

General idea

Find the sum of the smallest divisor and the largest divisor of each number in 1~n.

Ideas

Everyone knows that the smallest divisor of a number is $1.\; The$ largest divisor is itself.

That is equal to $\sum _{i=1}^n(1+i)$ , and then simplify:
$n+(1+n)\times n/2=2\times n/2+(1+n)\times n/2=(3+n)\times n/2$。

Then you can $O\left(1\right)$ asked.

Code

#include<cstdio>

using namespace std;

int n;

int main() {
    
    
	scanf("%d", &n);
	
	printf("%lld", 1ll * n * (n + 3) >> 1ll);
	
	return 0;
}