Midterm Question 3

1 like 1 dislike
6.5k 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 | 6.5k 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:172.69.214.204
©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 | 2.6k 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 | 11.6k 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 | 4.5k 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 | 4.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users