Midterm Question 3

1 like 1 dislike
7.3k 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 | 7.3k 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: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 | 3.1k 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 | 13k 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 | 5.2k 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 | 5.2k views
12,783 questions
183,442 answers
172,219 comments
4,824 users