Calculation 2(欧拉函数的运用)

Description

Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.

Input

For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.

Output

For each test case, you should print the sum module 1000000007 in a line.

Sample Input

3 4 0

Sample Output

0 2

运用欧拉函数求与n互质的数的总和：phi[n]*n/2;

证明：在1<x<n中，与n互质的数有{x1,x2,x3....,xn}，gcd(x,n)=gcd(n-x,n)=1,得：
{n-xn,....n-x2,n-x1},两个集合两两对应，相加除以2得phi[n]*n/2;

代码如下：

#include <stdio.h>
#include<math.h>
#include <string.h>

long long phi(long long n)
{
    long long res=n;
    for(int i=2; i*i<=n; i++)
    {
        if(n%i==0)
            {
                res=res-res/i;
                do
                    n/=i;
                while(n%i==0);
            }
    }
    if(n>1)
        res=res-res/n;
    return res;
}
int main()
{
    long long n,ans;
    while(scanf("%I64d",&n) != EOF)
    {
        if(n==0)
            break;
        ans=n*(n+1)/2-n;
        ans-=phi(n)*n/2;
        ans=ans%1000000007;
        printf("%I64d\n",ans);

    }

    return 0;
}