最大公约数问题

解法1：辗转相除法。

代码1：

int gcd(int a,int b)
{
	return (!b) ? a : gcd(b,a%b);
}

解法2：用减法替换除法。

代码2：

int gcdi(int a,int b)
{
	/*guarantee that a >= b*/
	if (a < b)
		return gcdi(b,a);
	if (b == 0)
		return a;
	else
		return gcdi(a - b, b);
}

解法3：利用奇偶性降低复杂度。

代码3：

int gcdii(int a,int b)
{
	/*guarantee that a >= b*/
	if (a < b)
		return gcdii(b,a);
	if (b == 0)
		return a;
	else {
		if (!(a & 0x1)){/* a is even */
			if (!(b & 0x1)) /* b is even*/
				return (gcdii(a >> 1,b >> 1) << 1);
			else /* b is odd */
				return (gcdii(a >> 1,b));
		}
		else { /* a is odd*/
			if (!(b & 0x1)) /* b is even */
				return gcdii(a,b >> 1);
			else /* b is odd */
				return gcdii(b,a - b);
		}
	}	
}

测试程序：

#include <stdio.h>
#include <assert.h>
#include <time.h>
#include <unistd.h>
#include <windows.h>
#include <stdlib.h>
#include <dos.h>

int gcd(int a,int b)
{
	return (!b) ? a : gcd(b,a%b);
}

int gcdi(int a,int b)
{
	/*guarantee that a >= b*/
	if (a < b)
		return gcdi(b,a);
	if (b == 0)
		return a;
	else
		return gcdi(a - b, b);
}

int gcdii(int a,int b)
{
	/*guarantee that a >= b*/
	if (a < b)
		return gcdii(b,a);
	if (b == 0)
		return a;
	else {
		if (!(a & 0x1)){/* a is even */
			if (!(b & 0x1)) /* b is even*/
				return (gcdii(a >> 1,b >> 1) << 1);
			else /* b is odd */
				return (gcdii(a >> 1,b));
		}
		else { /* a is odd*/
			if (!(b & 0x1)) /* b is even */
				return gcd(a,b >> 1);
			else /* b is odd */
				return gcd(b,a - b);
		}
	}	
}

int main()
{
	int a = 321232;
	int b = 329800;
	int r;
	
	clock_t s,t;
	s = clock();
	r = gcd(a,b);
	//Sleep(1000);
	t = clock();
	printf("%d\n",CLOCKS_PER_SEC);
	
	printf("Gcd(%d,%d) = %d . Time consumed %dms . \n",a,b,gcd(a,b),t-s);
	printf("Gcdi(%d,%d) = %d .\n",a,b,gcdi(a,b));
	printf("Gcdii(%d,%d) = %d .\n",a,b,gcdii(a,b));
}

REF：

1，编程之美 2.7