HDU - 1787 - GCD Again 【欧拉函数】题解

超、凢脫俗 2021-07-24 12:59 364阅读 0赞

### 目录 ###

*   *   *  1.题目
         *  2.代码

### 1.题目 ###

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

### 2.代码 ###

#include<iostream>
    #include<cstdio>
    #include<cmath>
    using namespace std;
    int euler(int n)
    { 
       int ans=n;
       for(int i=2;i<=sqrt(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(cin>>n)
    	{ 
    		if(n==0)  break;
    		cout<<n-euler(n)-1<<endl;
    	}
    	return 0;
    }