POJ 2407 Relatives（欧拉函数）

Relatives

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13044		Accepted: 6464

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

Source

Waterloo local 2002.07.01

 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。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int ac(int n)
{
	int res=n;
	int a=n;
	for(int i=2;i*i<=n;i++)
	{
		if(a%i==0)
		{
			res=res/i*(i-1);
			while(a%i==0)
			a/=i;
		}
	}
	if(a>1)
    res=res/a*(a-1);
	return res;
}
int main()
{
	int n;
	while(scanf("%d",&n),n)
	{
		int ans=ac(n);
		printf("%d\n",ans);
	}
	return 0;
}