Midterm Question 3

1 like 1 dislike
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 | 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:172.71.120.164
©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 | 3.4k 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 | 14.3k 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 | 5.6k 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 | 5.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users