用欧几里得算法实现两个数的最大公约数和最小公倍数

来源：互联网发布：与大津算法相似的算法编辑：程序博客网时间：2024/05/17 23:56

/************************************************************************
***By           : summon                                                                                                       ***
***Data      : 2010.01.09                                                                                                   ***
***Function : 求两个数的最大公约数和最小公倍数                                                                     ***
***                应用欧几里得算法计算gcd，用n*m=lcm*gcd计算lcm                                           ***
***Version : V1.0                                                                                                               ***
/************************************************************************/
#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef int STATUS;

STATUS InputData(int *Datan, int *Datam);
STATUS GCDofTwoNum(int Datan, int Datam, int *gcd);
STATUS LCMofTwoNum(int Datan, int Datam, int gcd, int *lcm);
STATUS OutputData(int gcd, int lcm);

int main(int argc, char* argv[])
{
int gcd;
int lcm;
int Datan;
int Datam;
int ret;

ret = InputData(&Datan, &Datam);
ret = GCDofTwoNum(Datan, Datam, &gcd);
ret = LCMofTwoNum(Datan, Datam, gcd, &lcm);
ret = OutputData(gcd, lcm);

return 0;
}

STATUS InputData(int *Datan, int *Datam)
{
if (Datam == NULL || Datan == NULL)
{
return 0;
}

while (1)
{
  printf("Please intput two positive int number:");
  fflush(stdin);
  scanf("%d %d", Datan, Datam);
  printf("/n");
  if (*Datan < 0 || *Datam < 0)
  {
   printf("You have inputted unlawful number, again!/n");
   continue;
  }else
  {
   break;
  }
}

return 1;
}

STATUS GCDofTwoNum(int Datan, int Datam, int *gcd)
{
if (gcd == NULL)
{
return 0;
}

    if (Datam > Datan)
{
  int temp = Datam;
  Datam = Datan;
  Datan = temp;
}else if (Datan == Datam)
{
  *gcd = Datam;
  return 1;
}

while (1)//循环体中n>m
{
  if (Datam == 0)
  {
   *gcd = Datan;
   return 1;
  }
  int ret = Datan%Datam;
  Datan = Datam;
  Datam = ret;
}
return 1;
}

STATUS LCMofTwoNum(int Datan, int Datam, int gcd, int *lcm)
{
if (lcm == NULL)
{
return 0;
}
*lcm = (int)(Datan*Datam)/gcd;
return 1;
}

STATUS OutputData(int gcd, int lcm)
{
printf("最大公约数=%d/n", gcd);
printf("最小公倍数=%d/n", lcm);
return 1;
}