Midterm Question 3

1 like 1 dislike
4.3k 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 | 4.3k 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:172.70.178.80
©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 | 1.8k 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 | 7.8k 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 | 3k 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 | 3.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users