Midterm Question 3

1 like 1 dislike
3.5k 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 | 3.5k 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:172.71.254.66
©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 | 1.4k 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 | 6.5k 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 | 2.6k 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 | 2.6k views
12,783 questions
183,443 answers
172,219 comments
4,824 users