Relatives（欧拉函数）

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

解题思路：欧拉函数：求不大于n且与n互质的数的个数。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int Euler(int x){
    int r=x;
    for(int i=2;i*i<=x;i++){//由于任何一个合数都至少有一个不大于根号x的质因子，所以只需遍历到根号x即可
        if(x%i==0){
            r=r/i*(i-1);//欧拉函数的通式，先除后乘，避免数据溢出
            while(x%i==0)x/=i;//消除x中所有质因子i，直到不能被i整除
        }
    }
    if(x>1)r=r/x*(x-1);//如果x大于1，说明还有一个质因子没有除掉
    return r;//返回个数
}
int main(){
    int n;
    while(cin>>n&&n)
        cout<<Euler(n)<<endl;
    return 0;
}