UVa 10299 :Relatives 欧拉函数

题目：
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
题目大意：输入n,求出1到n(包括n)中与n互质的整数的个数，注意当n=1时，结果为0

解题方法：欧拉函数

欧拉函数的定义：
欧拉函数是少于或等于n的数中与n互质的数的数目,计算公式为
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。

#include<iostream>//using namespace std;int euler(int n){//欧拉函数的实现    int ans=n,i;    if(n==1){        return ans==0;    }    for(i=2;i*i<=n;i++){        if(n%i==0){            ans=ans/i*(i-1);            while(n%i==0)                n /= i;         }    }    if(n>1)        ans=ans/n*(n-1);    return ans; } int main() {    int n;    while(cin>>n&&n!=0){        cout<<euler(n)<<endl;    }    return 0; }