hdoj 1787 GCD Again（欧拉函数）

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本文介绍了一道名为“GCDAgain”的ACM竞赛题目，该题要求计算与给定整数N的最大公约数大于1的所有整数M的数量（0<M<N）。文章提供了使用欧拉函数解决此问题的方法，并附带了C语言实现代码。

http://acm.hdu.edu.cn/showproblem.php?pid=1787

GCD Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2473 Accepted Submission(s): 1017

Problem Description

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

Sample Output

0
1

题目大意：输入一个数N，求2~N-1之间与N的最大公约数大于1的数的个数。

用欧几里得辗转相除法判断N和2~N-1之间的数的最大公约数会超时，要用欧拉函数。

#include<stdio.h>
#include<string.h>
int N;
int Euler(int n)
{
	int ret=n,i;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			ret=ret-ret/i;
			while(n%i==0)
			n=n/i;
		}
	}
	if(n>1) ret=ret-ret/n;
	return ret;
}
int main()
{
	while(scanf("%d",&N)&&N!=0)
	{
		printf("%d\n",N-1-Euler(N));
	}
	return 0;
}