poj 2407

        这是一道求欧拉函数的题目，题意十分直白，就是求欧拉函数的值，不过题目里关于互素的概念说的有点绕，让我看不太懂。

        欧拉函数φ(x)=x(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pn),其中p1, p2……pn为x的所有质因数，x是不为0的整数（摘自百度百科），所以找出x的素因子就可以求解了。这道题虽然n很大，不过直接算就好了，不会超时。

代码（C++）：

#include <cstdlib>
#include <iostream>
#include <cmath>
using namespace std;

int euler(int n)
{
    int i,tmp=n,len;
    len=sqrt(n*1.0);
    for(i=2;i<=len;i++)
    {
        if(n%i==0)
        {
           tmp=tmp/i*(i-1);
           while(n%i==0) n/=i;
        }
    }
    if(n>1) tmp=tmp/n*(n-1);
    return tmp;
}

int main(int argc, char *argv[])
{
    int n;
    while(cin>>n&&n!=0)
    {
        cout<<euler(n)<<endl;
    } 
    system("PAUSE");
    return EXIT_SUCCESS;
}

题目：

Relatives

Time Limit: 1000MS		Memory Limit: 65536K

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