Midterm Question 3

1 like 1 dislike
10.4k 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 | 10.4k 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:104.23.243.108
©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 | 4.1k 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 | 17.1k 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 | 6.7k 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 | 7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users