Midterm Question 3

1 like 1 dislike
11.6k 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 | 11.6k 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:104.23.243.108
©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 | 4.3k 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 | 18.2k 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 | 6.9k 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 | 7.3k views
12,783 questions
183,442 answers
172,219 comments
4,824 users