Midterm Question 3

1 like 1 dislike
8k views

Write a program that asks the user to enter two integers, then calculates and displays their greatest common divisor (GCD) and their Least common multiple (LCM).

寫一個輸入兩個整數 輸出它們的最大公因數程式

Hint: Use Euclidean Algorithm

使用歐幾里得演算法

LCM:

In arithmetic and number theory, the least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by LCM(a, b), is the smallest positive integer that is divisible by both a and b.

What is the LCM of 4 and 6?

Multiples of 4 are:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, ...

and the multiples of 6 are:

6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, ...

Common multiples of 4 and 6 are simply the numbers that are in both lists:

12, 24, 36, 48, 60, 72, ....

So, from this list of the first few common multiples of the numbers 4 and 6, their least common multiple is 12.

 

Example input: 

12 28

Example output (first number is GCD, second number is LCM):

4 84

 

[Exam] asked in Midterm by (12.1k points)
ID: 32707 - Available when: 2017-11-15 14:10 - Due to: Unlimited
edited by | 8k views

36 Answers

0 like 0 dislike 
Hidden content!
#include  * * **   *  
//gcd and lcm
int main(void)
{
***  *****    *  ********** ****  **  n,m,i,gcd, j,k,lcm;
   **  * *   *** ***** *     ****  ********   *   * *** &n, &m);
* *   ** *   *    ********  * i<=n * *    *   i<=m; i++)
* *****    * **** ***     * * ***
*  * **   * ***  ****  *** *****  *   *  *** * ******    **      ******* **    m%i==0)
*  *  **  **   *  **  ***  *      ****** *  * *   *  *   *    *   *******    *  **** ** 
* **  ****   ** **     * ****
   *****  *  *    * * **** * ***  **  ", gcd);

 * **  ** *   **  *  *        ** k>=n && k>=m; k--)
  *** *   **   ***    ** 
*  ** **   * ** ***  *** ** * ** *****     * *  * **  k%m==0)
* *  ** **  ** * *  ** *  **
  *      ****  *

  ***  * *  * * *** ****     *  *   ******* * * *   *  *


 *****       * *** ******  ** *   0;

}
answered by (-304 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>

int main() {
 *  ** * **  ****     * *    ** a,b,tmp,x,y;
  **   * *         ** **** *    *   %d", &x ,&y);
       *  ** * **  ******** 
  ** *  ****   *** *** **  *
 *  *   * **  **  * *   **  **     {
  ** ** *    ****  *** * **** **     * * ***   *  ** * * 
* * ** ** ***  *  ***** ** *      * *  *     ** **    **
* * * *         * **     ***   *  *  ***  *** * *** ** **  ** 
* **     ** **** *   ****  **
 * **   ** *** ***    * * *      **** %d", b, b*(x/b)*(y/b));
}
answered by (-120 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
int main(){
      * * ** ***   ** * *  *** **** a,b;
  *  ** *    **** *  **   *   gcd,lcm;
 *  *   * *   ** *  **  **   *****  * ** **  %d",&a,&b);

 **  * ****  * * *  **  * ag=a,bg=b;
 ** **  **  *     * **** *   **    *
*** ** *   **     **  **  * ** **** **   ******  **** *** 
 ***  **** *  *** ****   *    * ***   * ****  **    ****  temp;
* **  *  ***    *     ****  ** ** ** * * **   **  **      * * *  ** bg=ag, ag=temp;
***  * ***   *    **         

 *    *****      * ***  * i=1,al=a,bl=b;
  ***     *    *****   * ****   al%bl!=0 ) {
*     * *  **  * *   *     **  ***  *****  ******   * *
* ****   ****   *            *   *  ** *  *** *** ** *
 **   * *    * ** ***  * * ** 

 *** **** * * **  ** *  **** ***    *    *** %d",bg,al);
********   ** ** ** ** * * ** * 0;
}
answered by (-116 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
int a,b,i,q=1;
int main()
{
scanf("%d",&a);
scanf("%d",&b);

for(i=a;i>=1;i--)
{
    if(a%i==0)
 ****** * * * ***  * ******** ** **  *  **** ***     *  
*** *  **  ** * **  * *  ***** ***  *  ** * *  ***** ***   ****  **  **  * * ** * *** * * * *    *     **  * **** **  ***
 *  ***  *  ***  * ***   **** ** *   * **    *  * ** **  **** * * **** *** **     ** **** *     * *  *      ** **  **   * ******  ***  *** ** **
**  ** ***              *   * *** * * *  * ** *    *   *  ****  ***   * * *    *   * * * **** * ****  *  * *** ** ***    *** ***** ***  ***  
*    * **** **  * ** *** ** ** * * * *****     **** ** * *** *  *  ***    **** ****  *** ***   *  *   ** **  *  *** *   * **  *** *** ** *   ** **** * ** * ** * * ****** **
      **  *  **   **   ******  * ***   * ***  *  *  ** * *    ** **  **  **    * *** *** ** * * * *  ****    * **   *** *****  ***      * *    *    *     *    *  ",i);
 *** * ** ***   **  *   ***   * **** * *  * * *** *    ** *** * * * * * *  * **   ********  * ** ** **  **     **  **** ** * * *** ** *   ******* * *  **  *
 *  ******      * ****   ** * **   **  ** **    * **** **  ***  * *  ****   * ***  * ** **** * * ****** **   *    ** *   ******** ** * *** *
     ********    * ** ** * ****  ** ****** ** *   *  *
}


for(i=b;;i++)
{
    if(i%b==0&&i%a==0)
**   **  *****  *   *  * **** ** **  *  ** *   *   ** * **   
***  * *** *  * *  * *  * * *** **** *   *  ** **  *   ** ** ** ***  ****  ** 
*    *  **    ** *   ***  *  *  ***  *** *  * * *** 
        }

}




}
answered by (-336 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>\

int main()

{
    int a,b,i,j,k,flagGCD1,flagGCD2,flagLCM1,flagLCM2;
** ** ** *  ** **** ***  *  ** * *   * %d",&a,&b);

    //GCD
    for (i=a;i>0;i--)
    {
   ***    *  * **  *   **  ** ** **  *  ** ***  *   ******    
   * * **  **  *********  ** **    * ***   * *  **   ** *  * (flagGCD1==0)
* *  * **     ** **    ***  *** *   ****   * *     *  **  
*  ****     *  * *     * **   *** ***** ***  *   *  ***  ** *** *  ***     ** ***    *    ** 
 *** ** ** **     **    ***     *   ** *  *  * *******     * ** **   ** **   *   * (flagGCD2==0)
*** *****  * *  * * *** ** ******    *   ***    * ** * *** **  ** * * * **   *  
* *****  ** **      *****       * * *    *  ***    *       *      * ** ***** *  *   * **** * *   *  *    * **   **   *  ",i);
** * * **  **    * *   **** *   * * *** *  *** *    *  **** *   *    * **** ** ***           *** *      *   **
* *  *  *  ** **    ** * *  * *  ** *    *   *   ** * **  * *    *** **** **  **
  *   ****  *     **   * *      ** ** *  **   *      ** **
    }

    //LCM
    for (j=1;j;j++)
    {
 ** ** ** **  * **   * *   **  *    *  **  * **           *
***   * ********   * *   ****  *  *    ***  *    *** (k=1;k;k++)
  *    *   * ***** ** * * * ***      *  *  * **  *      *
  *** *  ** *   ** *    ******  ** *   * ****   *  *     *    *  * *  **   * *  *    **
*** * *   *** *  * * ***    * *  *  *** *  *****  * ***   * ** *       (flagLCM2==flagLCM1)
*  * **   ** **   *** **   * * ****** *   ** ** * *   **   ***** ****** *   * *  * *   *
**** **    ** * **     *  *   **** ** *    *   **   ****  *****  *   ** *   *  *   *  **  *  * *  ****    *  *  *****    ****       * * 
***  ****    *** * **  ******* **  * ** * *    *   **** ***      * * ** *    * *   * ** ** *  ** * * *   
*    ** *** *** * *   ** ***   * **  * *  *     *  **           ***  **     * ***
   **    * * * ** ***  **     **** **    * * *   ****** *   *** *  ** * ** (flagLCM2>flagLCM1)
  ** **     ** * ****     * * **   ** *  * * ********  **   **  *** * **   **  * ******   
 ***  ***  *  **  * *   ** * **  ****       ****  * *   ****  **  **  *   ***  ** ****   ***   * **      **  *
* * *****  *** ***** * * * **** *  **  *         *  ** ** * **  *  *** ** *  * *   *
 *****   ****  **    *      **   *    **    **** **  * *
   **   *    * *    ** **    *****  ******** **      *  **  (flagLCM2==flagLCM1)
**  *** **  *   * **** ** *  * ****    *** * **** 
* *   *** ***   *  *  *  *** *  *  *  * *    *  ** ****  * ********* * ** **  ** 
 * *  ***    * *** *   * *  *  *  ******   * *** *** *   ***
    }

    return 0;

}
answered by (-249 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
int a,b,i,q=1;
int main()
{
scanf("%d",&a);
scanf("%d",&b);

for(i=a;i>=1;i--)
{
    if(a%i==0)
* **     **    ***   ** ** * * *** ***   ** *        ***  * **
* * **  * *   *    * **  * * *  *  *** * * *  ***  *** **  ** * ****   *    ***  *   * * * *    *   **** **  *
*    * *   *        **  ***    *   * *  * **  **    *  *  * **  * *  ** *   *  **  * *  *   * *  * ***  * * *****   *  * *** * * * *** * ** ***    *** **     *
* **** ** ** * *****   **  *   * * **      * * *   * *  *** **  *** * * * ****    *   **  *  ****    *   ****    *  ****  ***  **  * *  ****   * **   * 
******* ***  **     ****  * *** **  *  *  **  *  * *  *  **    *   ** **  ** *****  *    ****     ********* *** * *    * **   *   ***        ***   * * * **** 
 *** *  * ***  *    ***  *  * * * * * ***  *** * ** *  *********  *  * *** * * ***  *    **  *  **** * * ** *** * *       * ***    *    **  *  *  ** * ** * *  *
* * * **  *** * ** * *** ***  ** ** *** ** *    * *  ****     * ***    * **  ******  ** ***  ** *  *      * * * * * ** ** ***** **  * ***  * *  ** *
  **   **** ****  * ********   *  *  ***    * ***  ***** **  *   **** **   ***** ** **** *** ** **  ***  ** ** ****  *****  ** *  ** ** **  * ***** *
*  *   *** ****  * *  * ** *     **   * *  * *   *
}


for(i=b;;i++)
{
    if(i%b==0&&i%a==0)
*   ***  ***  * *   * ** **  * *********  ** * **      
  * *  *  ***   ***      *   ****    *    *  ****  * * **     * **   *   *   *** 
*    ***  * *    *** **  *  **** * **  ****  * * 
        }

}




}
answered by (-336 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int main()
{
 * *  *            *   *** * *** a,b,smaller,i;

    *  ** *   *   * *   ** *  **  * ***    **  **  * ** *    *   

*     * *  ** *** *** *   *   **  ** *   
**  **  * * *  *  * * * **** *   ** *   * *** *    ** ** *** ** 
 ** *       *****  * ****  
 ** *   **   ** ** * *      *     ** *  * *     * *** * *    *  

 * * **  * * ** **   **  * * **   ** *   *** *
  * **     *** ****  * * *     *****  **   * **  *  ****   && b%i==0)
 ***** **** *** ******** *** **   * * *  ********  *  ** ** ******   *  * ***  *  **  

*  ** * *** *  *  **** ** **  * *****  * *   * **** ** ****** * ****  %d",i,(a/i)*b);

** * *  *  * *  *   *  * *   *   0;

}
answered by (-16 points)
edited by
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
commented by (4.3k points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
commented by (4.3k points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include<stdio.h>
int main()
{
****** *    ****     *  i,n1,n2,a,gcd,lcd;
**   *    **   *** ***   ********   **
*   *** *   *   *    * *** * **** ** * **  %d",&n1,&n2);
   *     * * ***    * *  **  *   *      || i<=n2; i++)
  *  **   * *****  * ***    * * 
  **   *** **   ***   *** * **  * *    * *     ** ** **  ** * * *  *  && n2%i==0)
*    **  ** * * ** ***  *  **** * *  * * *****  *    * 
  * *** * ****   *  * **  * *  * * *  ** * * * *** * *      * *** *  *  *** ***  **
**** *  ** *  **  *  **  * **** *    *   **** *     ** 
**  *****     ****  ** *  ** *
** * * **   **  * * *    * **  * ********* && a>=n2;a--)
    {
*** * * * *****    ** ** * **    **   ****  *  * ***   *    && a%n2==0)
** * *** ** * ** *  * ***  * *** * * *     *   *
***   ******  * ****** ***  *   *       * *  ***    *     ** **   *   * **    * ***  *  * 
 *  * **  * * *    *  *       *    ***   ** *  *  *  * 
 *  **  ***** *  ** * ****      
   *** ****  *** **   *  * *** *   * %d",gcd,lcd);
   **    ******  **  * * ***  * 0;
}
answered by (-323 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include           ***  
int main (){
int i,j,k;
***   ********* * * ***   *
 **  ***   * ** * **** **  * ** 
  *** * *** **
**   **** ***      ***  *  *    *   * ****** *    *  ** *  
*** *  ** ***** ** * *********
 **   **  * *** *  **   *


return 0;
}
answered by (90 points)
0 0
prog.c: In function 'main':
prog.c:6:10: error: expected expression before '=' token
 while(i!==0)||(j!==0){
          ^
commented by (4.3k points)
0 like 0 dislike 
Hidden content!
#include <stdio.h>
#include <stdlib.h>

int a,b,GDC,LCM;

int Get_GDC(int a1,int a2){

*   *** **  *   * *   *** i;
  *** *** * *  ** * ****  *** result=1;
**       **    *  *******    * **  *** *  *  *
       *** * ** * * *   ****  **   * *    ***  ** ***     *      *  * * && a2%i==0){
**  *  * **    *  *  ** * * ** * ** * *  ** * * *** * **    *** ** * * *     ** = i;
**   * ** **   *   ** * **  ***    * **   * **    ** 
  *     * * *   **   *  * 

    *   **   * **   *  * * *    result;
}
int Get_LCM(int a1,int a2){

  **    * **   ** *    *    *  i;
**  ********    ****   ** ****  result=1;
** ***   **** *** **  *  ***  ** **    * ***  * * 
 *** ** * *   * **** *   **  *******  *** * *    ***** *     *** * && i%a2==0){
***  * **    ********  **    * **  * * *** **     *** ********    ** *      ** * = i;
*     * * ** ** * *  ***  ** * * ** ** **   *   *     
     **   *     ***** ***

**     ***  * ****     * **** * result;
}


int main(){

** * **    *   * ** *  *   **** *     * *   *** * * * * ***  ****** 

 ** * **** ***  *   *    * ** *  ** * *
* *       **    * *****   ** *** *   * ** 



    * *  * ****    *** ** **  %d", GDC, LCM);
*   *     *   **** * ** *   0;
}
answered by (-286 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
commented by (4.3k points)
Page:
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:108.162.241.184
©2016-2025

Related questions

0 like 0 dislike
10 answers
Midterm Question 3 - Makeup
[Exam] asked Dec 9, 2017 in Midterm by thopd (12.1k points)
ID: 36752 - Available when: 2017-11-15 14:10 - Due to: Unlimited | 3.4k views
0 like 0 dislike
69 answers
10-26 Exercise 1: Calculates and displays the GCD of 2 integers
[Exercise] Coding (C) - asked Oct 26, 2017 in Chapter 6: Loops by semicolon (5.2k points)
ID: 28913 - Available when: 2017-10-26 18:00 - Due to: Unlimited | 14.3k views
0 like 0 dislike
21 answers
Midterm Question 5 - Makeup
[Exam] asked Dec 9, 2017 in Midterm by thopd (12.1k points)
ID: 36754 - Available when: 2017-11-15 14:10 - Due to: Unlimited | 5.6k views
0 like 0 dislike
22 answers
Midterm Question 4 - Makeup
[Exam] asked Dec 9, 2017 in Midterm by thopd (12.1k points)
ID: 36753 - Available when: 2017-11-15 14:10 - Due to: Unlimited | 5.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users