Midterm Question 3

1 like 1 dislike
4.3k 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 | 4.3k 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:172.70.100.168
©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 | 1.8k 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 | 7.8k 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 | 3k 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 | 3.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users