Midterm Question 3

1 like 1 dislike
9.2k 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 | 9.2k 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:162.159.115.32
©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 | 3.7k 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 | 15.5k 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 | 6.2k 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 | 6.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users