Midterm Question 3

1 like 1 dislike
1.2k 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 | 1.2k 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:108.162.216.107
©2016-2024

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 | 511 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 | 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 | 961 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 | 970 views
12,783 questions
183,443 answers
172,219 comments
4,824 users