Midterm Question 3

1 like 1 dislike
7.5k 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 | 7.5k 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:172.69.130.52
©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 | 3.2k 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 | 13.4k 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 | 5.3k 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 | 5.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users