【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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you should print the sum module 1000000007 in a line.

[align=left]Sample Input[/align]

3
4
0

[align=left]Sample Output[/align]

0
2

[align=left]Author[/align]
GTmac

[align=left]Source[/align]
2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

[align=left]Recommend[/align]
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