学步园

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 contain several test case.
A number N per line.

For each N, output ,∑gcd(i, N) 1<=i <=N, a line

2
6

3
15

POJ Contest,Author:Mathematica@ZSU

题意：求 ∑gcd(i, N) 1<=i <=N

题解：涉及欧拉函数，积极函数，莫比乌斯反演等...本弱菜无法说明，搞了很久，但是代码非常短的说.....这大概就是典型的数论题的特征吧....

#include<stdio.h>
int main()
{
    long long n,temp,res,i,cou;

    while(scanf("%I64d",&n)>0)
    {
        for(res=1,i=2;i*i<=n;i++)
        {
            if(n%i!=0) continue;
            cou=0;  temp=1;
            while(n%i==0){ n/=i; temp*=i; cou++; }
            res*=temp/i*(i-1)*cou+temp;
        }
        if(n>1) res*=2*n-1;
        printf("%I64d\n",res);
    }

    return 0;
}