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Fun with Integers CodeForces - 1062D

​​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;
}

 


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