Midterm Question 3

1 like 1 dislike
9.6k 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 | 9.6k 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:104.23.197.65
©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 | 3.9k 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 | 16.2k 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 | 6.4k 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 | 6.6k views
12,783 questions
183,442 answers
172,219 comments
4,824 users