hdu 3501

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

340

 

Sample Output

02

 

Author

GTmac

 

Source

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

 

Recommend

zhouzeyong   |   We have carefully selected several similar problems for you:  3507 3506 3505 3504 3499 

欧拉函数

 

     对正整数n，欧拉函数是少于或等于n的数中与n互质的数的数目。例如euler(8)=4，因为1,3,5,7均和8互质。
     Euler函数表达通式：euler(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…(1-1/pn),其中p1,p2……pn为x的所有素因数，x是不为0的整数。euler(1)=1（唯一和1互质的数就是1本身）。
     欧拉公式的延伸：一个数的所有质因子之和是euler(n)*n/2。
     欧拉函数的实现也非常的简单，在原有的素因子分解基础上，利用一下迭代就好了。

     这个题就是前n-1项的和再减去euler(n)*n/2,结果在求个模就行了

#include <iostream>using namespace std;long long euler(long long n){  long long res=n,a=n;  for(int i=2;i*i<=a;i++)  {   if(a%i==0)   {    res=res/i*(i-1);//先进行除法是为了防止数据的溢出    while(a%i==0) a/=i;   }  }  if(a>1) res=res/a*(a-1);  return res;}int main(){    long long n,num,ans;    while(cin>>n&&n!=0)    {     num=euler(n);     num=n-1-num;     ans=num*n/2;     cout<<ans%1000000007<<endl;    }    return 0;}