Midterm Question 3

1 like 1 dislike
10.8k 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 | 10.8k 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:104.23.197.94
©2016-2026

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 | 4.2k 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 | 17.5k 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 | 6.8k 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 | 7.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users