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:104.23.190.141
©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 | 11.7k 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 | 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