GDUFE ACM-1093

题目：http://acm.gdufe.edu.cn/Problem/read/id/1093

 

互质数简单版

Time Limit: 2000/1000ms (Java/Others)

Problem Description:

  For given integer N (1<=N<=10^5) find amount of positive numbers not greater than N that coprime with N. Let us call two positive integers (say, A and B, for example) coprime if (and only if) their greatest common divisor is 1. (i.e. A and B are coprime iff gcd(A,B) = 1). 

Input:

The input includes several cases. For each case,the input contains integer N. 

Output:

For each case,print answer in one line. 

Sample Input:

9

Sample Output:

6

思路：一个个的判断是不是互质数，如果是，总数加1，这个方法数字大的时候回超时==

难度：简单吧

代码：

#include<stdio.h>
int main()
{
    int n,i,c,sum;
    int gcd(int a,int b);
    while(~scanf("%d",&n))
    {
        sum=0;
        for(i=1;i<n;i++)
        {
            c=gcd(n,i);
            if(c==1)
                sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}

int gcd(int a,int b)
{
    int c;
    while(a%b!=0)
    {
        c=a%b;
        a=b;
        b=c;
    }
    return b;
}