Midterm Question 3

1 like 1 dislike
8.5k 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 | 8.5k 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:108.162.241.97
©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 | 3.6k 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 | 14.9k 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 | 5.8k 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 | 6k views
12,783 questions
183,442 answers
172,219 comments
4,824 users