http://codeforces.com/contest/1062/problem/D
考虑每个乘数x的贡献
x=2时 (+-2,+-4) (+-3,+-6) (+-4,+-8)...
x=3时 (+-2,+-6) (+-3,+-9) (+-4,+-12)...
......
规律就很显然了
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
int main()
{
ll ans,n,i;
scanf("%lld",&n);
ans=0;
for(i=2;i<=n;i++){
ans+=(4ll*(n/i-1))*i;
}
printf("%lld\n",ans);
return 0;
}