Midterm Question 3

1 like 1 dislike
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 | 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:172.69.6.98
©2016-2024

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 | 486 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 | 1.9k 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 | 917 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 | 919 views
12,783 questions
183,443 answers
172,219 comments
4,824 users