Midterm Question 3

1 like 1 dislike
5.4k 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 | 5.4k 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:172.69.59.92
©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 | 2.2k 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 | 9.5k 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 | 3.7k 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 | 3.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users