Midterm Question 3

1 like 1 dislike
9.2k 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 | 9.2k 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:162.159.115.32
©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 | 3.7k 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 | 15.5k 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 | 6.2k 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 | 6.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users