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:172.70.100.193
©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 | 6.4k 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