hdu 3501 数论 与n不互质的数的和

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1241    Accepted Submission(s): 518

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
   
   
   
   

 

Author

GTmac

 

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

题意：

输入n  计算比n小的数中 和n不互质的数的和   %1000000007

 

/*   if  gcd(n,i)==1     then gcd(n,n-i)==1
    所以 那么对于n有phi(n)(欧拉函数)个小于n的数与n互质 
	由上面的公式可以知道 其中一个若为i则存在一个为n-i
	那么2者之和为n  这样的一共有phi(n)/2对 
	故与n互质的所有数的和为 n*phi(n)/2
	那么与n不互质的 数就是(1+n-1)*(n-1)/2-n*phi(n)/2
*/

#include<stdio.h>
#include<math.h>
#define mod 1000000007
__int64 get_phi(__int64 x)// 就是公式   
{  
    __int64 i, res=x;  
    for (i = 2; i <(__int64)sqrt(x * 1.0) + 1; i++)  
        if(x%i==0)  
        {  
            res = res / (__int64)i * (i - 1);  
            while (x % i == 0) x /= i; // 保证i一定是素数    
        }  
        if (x > 1) res = res / (__int64)x * (x - 1);//这里小心别溢出了   
        return res;  
} 
int main()
{
	__int64 n,ans;
	while(scanf("%I64d",&n)!=EOF)
	{
		if(!n) break;
		ans=((__int64)(1+n-1)*(n-1)/2-(__int64)n*get_phi(n)/2)%mod;//不能先除2再乘
		printf("%I64d\n",ans%mod);
	}
	return 0;
}