poj2480 Longge's problem

Description

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

26

Sample Output

315

设函数g(n) = gcd(i,n) (1<=i<=n)，由积性函数的定义，g(n)=g(m1)*g(m2) (n=m1*m2 且 （m1, m2）= 1)，所以g是积性函数。由具体数学上的结论，积性函数的和也是积性的。所以f(n) = ∑gcd(i, n)也是积性函数。由初等数论中的定理，如果f(n)是不恒为0的数论函数，n>1时 n=p1^a1*p2^a2*...*ps^as，那么f(n)是积性函数的充要条件是f(1)=1，及f(n) = f(p1^a1)*f(p2^a2)*...f(pr^ar)。所以只要求f(pi^ai)就好，如果d是n的一个约数，那么1<=i<=n中gcd(i,n) = d的个数是phi(n/d)，即n/d的欧拉函数

f(pi^ai) =  Φ（pi^ai）+pi*Φ（pi^(ai-1)）+pi^2*Φ（pi^(ai-2)）+...+pi^(ai-1)* Φ（pi）+ pi^ai *Φ（1）

     = pi^(ai-1)*(pi-1) + pi*pi^(ai-2)*(pi-1)....+pi^ai

     =  pi^ai*(1+ai*(1-1/pi))

f(n) = p1^a1*p2^a2...*pr^ar*(1+a1*(1-1/p1))*(1+a2*(1-1/p2))*...

       =  n*(1+a1*(1-1/p1))*(1+a2*(1-1/p2))*...

#include<stdio.h>#include<string>#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <vector>#include <queue>#include <cstdio>#include <cmath>#include <string>#include <stack>#include <cctype>using namespace std;int main(){    __int64 n;    while(~scanf("%I64d",&n))    {        __int64 i,j,k,l,ans,v;        ans=n;        v=sqrt(n);        for(i=2; i<=v; i++)        {            if(n%i==0)            {                k=0;                l=i;                while(n%l==0)                {                    k++;                    n/=l;                }                ans+=ans*k*(l-1)/l;            }        }        if (n!=1)        {            ans = ans + ans*(n-1)/n;        }        printf("%I64d\n", ans);    }    return 0;}