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
分析：
求一个数的所有的小于该数且与该数互质的数的个数，也就是欧拉函数的应用。
直接套欧拉函数公式，即将n素分解后有n=p1k1*p2k2…pm^km，则euler(n)=n(1-1/p1)(1-1/p2)…(1-1/pm) 。
代码：
#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
ll get_euler(ll n)//欧拉函数的应用
{
ll ans=n;
for(ll 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()
{
ll n;
while(scanf("%lld",&n),n)
printf("%lld\n",get_euler(n));
return 0;
}