【hdu - 3501 (数论、欧拉函数)】

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1121    Accepted Submission(s): 479

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

 

 

Recommend

zhouzeyong

 

 

 

// Project name : 3501 ( Calculation 2 ) 
// File name    : main.cpp
// Author       : Izumu
// Date & Time  : Fri Jul 13 17:14:43 2012


#include <iostream>
#include <stdio.h>
#include <string>
#include <math.h>
#include <algorithm>
using namespace std;

////////////////////////////////////////////////////////////////////////
typedef unsigned long long int longint;
////////////////////////////////////////////////////////////////////////
longint phi(longint num)
{
    longint sum = 1;
    for (longint i = 2; i <= sqrt((double long)num); i++)
    {
        if (num % i == 0)
        {
            while (num % i == 0)
            {
                sum *= i;
                num /= i;
            }
            sum /= i;
            sum *= (i - 1);
        }
    }
    if (num != 1)
    {
        sum *= (num - 1);
    }
    return sum;
}
////////////////////////////////////////////////////////////////////////
int main()
{
    longint number;
    while (cin >> number && number)
    {
        cout << ((number - 1) * number / 2 - number * phi(number) / 2) % 1000000007 << endl;
    }
    return 0;
}

// end 
// ism