Midterm Question 3

1 like 1 dislike
1.3k 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 | 1.3k 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:172.69.58.178
©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 | 515 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 | 2.3k 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 | 1.1k 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 | 1k views
12,783 questions
183,443 answers
172,219 comments
4,824 users