hdu3501(数学)

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1650    Accepted Submission(s): 697

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    求小于n的不与n互质的数的和，可以用欧拉函数求得这样的数个数
   
   
   
   
   
   
   
   

    
    
    
    有数学理论：grd(n,i)=1,则grd(n,n-i)=1,所以与n互质的数的和为phi(n)*n/2;
   
   
   
   
   
   
   
   

    
    
    
    综上,不与n互质且小于n的数的和为n*(n+1)/2-phi(n)*n/2-n;
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const __int64 mod=1000000007;

__int64 phi(__int64 x)  
{  
    __int64 i,res = x;  
    for(i=2;i<(__int64)sqrt(x*1.0)+1;i++)  
    {  
        if(x%i==0)  
        {  
           res = res / i * (i-1);  
            while(x % i == 0)  
                x /= i;  
        }  
    }  
    if(x > 1)  
        res = res / x * (x - 1);  
    return res;  
}

int main()
{
	__int64 n;
	__int64 ans;
	while(~scanf("%I64d",&n))
	{
		if(0==n)
			break;
		//cout<<phi(n)<<endl;
		printf("%I64d\n",phi(n));
		ans=((n-1-phi(n))*n/2);
		printf("%I64d\n",ans%mod);

	}
	return 0;
}