Midterm Question 3

1 like 1 dislike
6.5k 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 | 6.5k 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:172.70.80.170
©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 | 2.6k 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 | 11.7k 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 | 4.5k 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 | 4.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users