GCD Again HDU 1787（欧拉函数水题）_gcd again hdu

本文链接：https://blog.youkuaiyun.com/codeswarrior/article/details/81533478

本文介绍了一个基于欧拉函数解决的问题——寻找小于N的所有正整数M的数量，使得N与M的最大公约数大于1。该问题是对传统GCD问题的一个拓展，并提供了一段简洁的C++代码实现。

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GCD Again

 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

直接用欧拉函数直接求

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int Euler(int n){
    int ans = n;
    for(int 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(scanf("%d",&n) != EOF){
        if(!n) break;
        printf("%d\n",n-Euler(n)-1);//减去n本身
    }
    return 0;
}