Midterm Question 3

1 like 1 dislike
8.3k 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 | 8.3k 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:172.69.59.138
©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 | 3.5k 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 | 14.7k 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 | 5.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users