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
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.
Output
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.
Sample Input
2
4
0
Sample Output
0
1

题意：给出一个n计算出一个M(0<M<N)，并且n和M的最大公约数大于1

思路：就是让求小于n且不与n互质的数的个数。算出小于n且与n互质的数的个数（欧拉函数的值）然后用n减去这个值，注意这里M不能等于n，所以最后结果要减去1.（注意：这里的数据量太大，开不了这么大的数组，所以虽然是多组数据也不能预处理）

AC代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int phi(int x)
{
    int ans = x;
    for(int i = 2; i*i <= x; i++)
    {
        if(x % i == 0)
        {
            ans = ans / i * (i-1);
            while(x % i == 0) x /= i;
        }
    }
    if(x > 1) ans = ans / x * (x-1);
    return ans;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n)
    {
         int t;
         t=phi(n);
        printf("%d\n",n-t-1);
    }
}