poj2480Longge's problem数论积性函数

Longge’s problem

Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N.
“Oh, I know, I know!” Longge shouts! But do you know? Please solve it.
Input
Input contain several test case.
A number N per line.
Output
For each N, output ,∑gcd(i, N) 1<=i <=N, a line
Sample Input
2
6
Sample Output
3
15

数论还是难呀，看了一下午终于看懂了。
官方解析如下：

但是，依旧看不懂AC代码。
我们通过一系列证明：（不会Latex的弱鸡）

代码如下：

#include <iostream>
using namespace std;
typedef long long ll;
int main()
{
    ll n;
    while(cin>>n){
        ll ans=n;
        for(ll i=2;i*i<=n;i++){
            if(n%i==0){
                ll a=0,p=i;
                while(n%p==0){
                    a++;
                    n/=p;
                }
                ans=ans+ans*a*(p-1)/p;
            }
        }
        if(n!=1)
            ans=ans+ans*(n-1)/n;
        cout<<ans<<endl;
    }
    return 0;
}