求GCD(最大公约数)的算法

int GCD ( int a , int b) //递归版
{
    if ( b != 0 )  return GCD( b, a%b );
    return a;
}
int GCD ( int a , int b) //非递归版
{
    int c;
    while (b)
    {
        c = a;
        a = b;
        b = c % a;
    }
    return a;
}

另一种方法:

若x,y均为偶数,gcd(x,y) = 2 * gcd(x/2,y/2);

若只x均为偶数,gcd(x,y) = gcd(x/2,y);

若只y均为偶数,gcd(x,y) =  gcd(x,y/2);

若x,y均为奇数,gcd(x,y) = gcd(y, x- y);(两个奇数相减,必得偶数)
 

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转载自blog.csdn.net/zhouchenghao123/article/details/83995306