Midterm Question 3

1 like 1 dislike
9k 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 | 9k 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:104.23.197.64
©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 | 15.4k 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 | 6.1k 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 | 6.3k views
12,783 questions
183,442 answers
172,219 comments
4,824 users