Euler Numbers

/*Description

Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz. Input

There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case. Output

For each test case there should be single line of output answering the question posed above. Sample Input

7 12 0

Sample Output

6 4

#include <iostream>
#include<cmath>
#include <algorithm>
using namespace std;
int Euler(int n)
{
	int ret=n,p;
	for( p=2;p*p<=n;p++)
	{
		if(n%p==0)
		{
			ret=ret/p*(p-1);
			while(n%p==0)
				n/=p;
		}
	}
	if(n>1)
		ret=ret/n*(n-1);
  //	注意与后面的比较，时间更快，只有一个比sqrt(n)大的一个素数因子
return ret;
}
int main()
{
	int n;
	while(scanf("%d",&n),n)
		printf("%d\n",Euler(n));
	return 0;
}

#include <iostream>
#include<cmath>
#include <algorithm>
using namespace std;
int Euler(int n)
{
	int ret=n,p;
	for( p=2;p<=n;p++)
	{
		if(n%p==0)
		{
			ret=ret/p*(p-1);
			while(n%p==0)
				n/=p;
		}
	}
	if(n>1)
		ret=ret/p*(p-1);
	return ret;
}
int main()
{
	int n;
	while(scanf("%d",&n),n)
		printf("%d\n",Euler(n));
	return 0;
}