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.

lcy

2007省赛集训队练习赛（10）_以此感谢DOOMIII 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;

const int n=7000000;

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