Midterm Question 3

1 like 1 dislike
10.1k 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 | 10.1k 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:162.159.115.10
©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 | 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 | 16.7k 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 | 6.6k 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 | 6.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users