Midterm Question 3

1 like 1 dislike
5k 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 | 5k 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:172.70.126.94
©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 | 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 | 8.8k 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 | 3.4k 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 | 3.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users