Midterm Question 3

1 like 1 dislike
6.8k 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 | 6.8k 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:172.69.17.66
©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 | 2.7k 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 | 11.9k 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 | 4.7k 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 | 4.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users