Euclidean algorithm

// tmp.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
using namespace std;

int gcd(int num1, int num2)
{
    // Any of each is 0, the result must be zero.
    if (0 == num1 || 0 == num2)
    {
        return 0;
    }//if

    // This is based on a fact that
        // if
        //    A % x = 0 and B % x = 0
        // then
        //      ( A - B ) % x = 0
        // so, we just reduce the scope to get the result
        // step by step.
    int remainder = num2;
    int remainder1 = 0;    // num1 % remainder: reduce once redundant evaluated
    int temp = 0;          // Reduce once redundant evaluated
    int cnt = 0;           // Counter how many times
                                // arithmetics have been execute.
    while (temp = 0 == (remainder1 = num1 % remainder) ?
                            num2 % remainder : remainder1)
    {
        remainder = temp;

        cnt++;
    }//while

    cout <<cnt <<" times' arithmetics executed" <<endl;
    return remainder;
}

int gcd_classic(int a, int b)
{
    int c = a < b ? a : b;        // The greatest common divisor.
    int tmp = 0;                // Reduce reduplicated evaluation of 'a % b'.
    int cnt = 0;                // Counter how many times
                                    // arithmetics have been execute.

    while (0 != (tmp = a % b))
    {
        c = tmp;

        a = b;
        b = c;

        cnt++;
    }

    cout <<cnt <<" times' arithmetics executed" <<endl;
    return c;
}

int gcd_error(int num1, int num2);

int main()
{
    // 经典的'辗转相除法'语法简洁, 效率很高.
        // 我尝试了一个自己的算法, 暂且称为'缩减算法', 是
        // 看了'辗转相除法'后想尝试一种更好记忆的算法而产生的.
        // 在尝试的过程中还因为一些错误误以为自己发现了更好的算法.
        // 想来这一下午没白费, 至少对'辗转'有了更深刻的印象和理解.
        // 之前错误的算法我列在最后了. 以供参考.
    cout <<"Result: " <<gcd(1448442, 10024) <<endl;
    cout <<"Result: " <<gcd_classic(1448442, 10024) <<endl;
    cout <<"Error Result: " <<gcd_error(1448442, 10024) <<endl;

    return 0;
}

int gcd_error(int num1, int num2)
{
    // Any of each is 0, the result must be zero.
    if (0 == num1 || 0 == num2)
    {
        return 0;
    }//if

    int remainder = num2;
    int cnt = 0;           // Counter how many times
                                // arithmetics have been execute
    while (0 != num1 % remainder)   // Error occur here!
    {
        remainder = num1 % remainder;
        cnt++;
    }//while

    cout <<cnt <<" times' arithmetics executed" <<endl;
    return remainder;
}