Midterm Question 3

1 like 1 dislike
9.6k 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 | 9.6k 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:104.23.197.65
©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 | 3.9k 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 | 16.2k 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 | 6.4k 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 | 6.6k views
12,783 questions
183,442 answers
172,219 comments
4,824 users