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:172.70.50.51
©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