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=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            res=res/i*(i-1);
            while(x%i==0)
               x=x/i;
        }
    }
    if(x>1)
        res=res/x*(x-1);
    return res;
}

int main()
{
    ll n,sum;
    while(scanf("%lld",&n)!=EOF)
    {
        sum=0;
        ll i;
        for(i=1;i*i<=n;i++)
        {
            if(n%i==0)
                sum+=i*euler(n/i)+(n/i)*euler(i);
        }
        i--;
        if(i*i==n)
            sum-=i*euler(i);
        printf("%lld
",sum);
    }
    return 0;
}