问题
实现欧几里得算法来查找两个整数的最大公约数 (GCD) 和最小公倍数 (LCM),并将结果与给定整数一起输出。
解决方案
实现欧几里得算法求两个整数的最大公约数 (GCD) 和最小公倍数 (LCM) 的解决方案如下 -
求 GCD 和 LCM 的逻辑如下 -if(firstno*secondno!=0){
gcd=gcd_rec(firstno,secondno);
printf("<p>The GCD of %d and %d is %d</p><p>",firstno,secondno,gcd);
printf("</p><p>The LCM of %d and %d is %d</p><p>",firstno,secondno,(firstno*secondno)/gcd);
}</p>
调用的函数如下 -
int gcd_rec(int x, int y){
if (y == 0)
return x;
return gcd_rec(y, x % y);
}
程序
以下是 C 程序,用于实现欧几里得算法,以求两个整数的最大公约数 (GCD) 和最小公倍数 (LCM) -
现场演示
#include<stdio.h>
int gcd_rec(int,int);
void main(){
int firstno,secondno,gcd;
printf("Enter the two no.s to find GCD and LCM:");
scanf("%d%d",&firstno,&secondno);
if(firstno*secondno!=0){
gcd=gcd_rec(firstno,secondno);
printf("<p>The GCD of %d and %d is %d</p><p>",firstno,secondno,gcd);
printf("</p><p>The LCM of %d and %d is %d</p><p>",firstno,secondno,(firstno*secondno)/gcd);
}
else
printf("One of the entered no. is zero:Quitting</p><p>");
}
/*Function for Euclid's Procedure*/
int gcd_rec(int x, int y){
if (y == 0)
return x;
return gcd_rec(y, x % y);
}</p>
输出
当执行上述程序时,会产生以下结果 -
Enter the two no.s to find GCD and LCM:4 8
The GCD of 4 and 8 is 4
The LCM of 4 and 8 is 8