hdoj 3501 【欧拉函数 求小于或者等于n的数中 与n互质的数总和】

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2548    Accepted Submission(s): 1064

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    题目意思： 求小于或者等于n的数中，与n互质的数之和。
   
   
   
   
   
   
   
   

    
    
    
    欧拉函数扩展： 小于或者等于n的数中(n > 1)，与n互质的数总和 -> euler[n] * n / 2。
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    代码：
   
   
   
   
   
   
   
   

    
    
    
    
//思路：欧拉函数的引伸：小于或等于n的数中，与n互质的数的总和为：euler(x) * x / 2。(n>1)
#include <cstdio>
#include <cstring>
#include <cmath>
#define LL long long 
#define MOD 1000000007
using namespace std;
LL euler(LL n)
{
	LL i;
	LL eu = n;
	for(i = 2; i*i <= n; i++)
	{
		if(n % i == 0)
		{
			eu = eu * (i-1) / i;
			while(n % i == 0)
			n /= i;
		}
	}
	if(n > 1) eu = eu * (n-1) / n;
	return eu;
}
int main()
{
	LL n;
	while(scanf("%lld", &n), n)
	{
		LL ans = n * (n+1) / 2;
		ans = ans - euler(n) * n / 2;
		printf("%lld\n", (ans-n)%MOD);
	}
	return 0;
}