编程之美--2.7

题目描述：最大公约数问题

思路：三个思路

(1)gcd(x,y) = gcd(y,x%y);

(2)gcd(x,y) = gcd(y,x-y);

(3)gcd(x,y) = gcd(x/2,y/2)*2;(x,y均被2整除)

#include <iostream>
#include <queue>
#include <climits>
#include <algorithm>
#include <memory.h>
#include <stdio.h>
using namespace std;

static int num_1 = 0;
static int num_2 = 0;
static int num_3 = 0;
int gcd_1(int a,int b)
{
    if(b == 0)
    {
        ++num_1;
        return a;
    }
    else
    {
        ++num_1;
        return gcd_1(b,a%b);
    }
}
int gcd_2(int a,int b)
{
    if(b == 0)
    {
        ++num_2;
        return a;
    }
    if(a < b)
    {
        ++num_2;
        return gcd_2(b,a);
    }
    ++num_2;
    return gcd_2(b,a-b);
}
int gcd_3(int a,int b)
{
    ++num_3;
    if(b == 0)
    {
        return a;
    }
    if(a < b)
    {
        return gcd_3(b,a);
    }
    if(a % 2 == 0)
    {
        if(b % 2 == 0)
        {
            return (gcd_3(a>>1,b>>1)<<1);
        }
        else
        {
            return gcd_3(a>>1,b);
        }
    }
    else
    {
        if(b % 2 == 0)
        {
            return gcd_3(a,b>>1);
        }
        else
        {
            return gcd_3(b,a-b);
        }
    }
}

int main()
{
    int a = 12;
    int b = 15;
    cout<<gcd_1(a,b)<<" "<<gcd_2(a,b)<<" "<<gcd_3(a,b)<<endl;
    cout<<num_1<<" "<<num_2<<" "<<num_3<<endl;
    return 0;
}