Midterm Question 3

1 like 1 dislike
7.3k 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 | 7.3k 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:172.71.194.142
©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 | 3.1k 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 | 13k 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 | 5.2k 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 | 5.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users