Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5342    Accepted Submission(s): 2199

Problem 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

題意：求解1~n與n不互質的數的和。

解析：典型的尤拉函式應用。

利用尤拉函式即可求解，1~n比n小且與n互素的數的總和為
*        sum(n) = n * phi(n) / 2;那麼可以先求出1~n-1的總和，然後
*        減去sum(n)即可。
 

#include<bits/stdc++.h>
using namespace std;
 
#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}

ull euler(ull n)
{
    ull res=n,a=n;
     for(ull i=2;i*i<=a;i++)
	 {
        if(a%i==0)
		{
	        res=res/i*(i-1);
	        while(a%i==0) a/=i;
        }
     } 
     if(a>1) res=res/a*(a-1);  
     return res;
}

int main()
{
	ull n;
	while(~scanf("%llu",&n),n)
	{
		ull sum=n*(1+n)/2-n,ans;
		ans=sum-euler(n)*n/2;
		printf("%llu\n",ans%mod);
	}
	return 0;
}