poj2480-Longge's problem-(欧拉函数)

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
翻译：给一个数n,1<=i<=n,求gcd(i,n)之和。
解题过程：
令d=gcd(i,n)，d必然是n的因子，并且是i和n的最大公因子。
则gcd(i/d,n/d)=1。令y=n/d,通过y求出与y互质的数的个数，然后对这个数量乘以最大公因子d，累加。

 #include <iostream>
 #include<stdio.h>
 #include <algorithm>
 #include<string.h>
 #include<cstring>
 #include<math.h>
 #define inf 0x3f3f3f3f
 #define ll long long
 using namespace std;

 ll euler(ll x)
 {
     ll res=x;
     for(ll i=;i*i<=x;i++)
     {
         if(x%i==)
         {
             res=res/i*(i-);
             while(x%i==)
                x=x/i;
         }
     }
     if(x>)
         res=res/x*(x-);
     return res;
 }

 int main()
 {
     ll n,sum;
     while(scanf("%lld",&n)!=EOF)
     {
         sum=;
         ll i;
         for(i=;i*i<=n;i++)
         {
             if(n%i==)
                 sum+=i*euler(n/i)+(n/i)*euler(i);
         }
         i--;
         if(i*i==n)
             sum-=i*euler(i);
         printf("%lld\n",sum);
     }
     return ;
 }