Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The greatest common divisor GCD (a, b) of two positive integers a and b, sometimes written (a, b), is the largest divisor common to a and b. For example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the Euclidean algorithm. Now I am considering a little more difficult problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
Good Luck!

Input contains multiple test cases. Each test case contains an integers N (1<N<100000000). A test case containing 0 terminates the input and this test case is not to be processed.

For each integers N you should output the number of integers M in one line, and with one line of output for each line in input.

2
4
0

0
1

lcy

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题意：

求小于n的gcd(i,n)大于1的个数 

思路 ： 欧拉函数直接求gcd(i,n)==1的个数  用n减即可，注意小于n，故再减去1.

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>

using namespace std;


int p(int n)
{
	int ans=n;
	for(int i=2;i*i<=n;i++){
		if(n%i==0){
			ans-=ans/i;
			while(n%i==0){
				n/=i;
			}
		}
	}
	if(n>1)ans-=ans/n;
	return ans;
}
int main()
{
	int n;
	while(scanf("%d",&n),n)
	printf("%d\n",n-p(n)-1);
    return 0;
}