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 multiplelowest common multiple, or smallest common multiple of two integers a and b, usually denoted by LCM(ab), 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
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
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
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
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
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
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
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
Case 4: Wrong output
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct output
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
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){
          ^
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
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
[Exam] asked Dec 9, 2017 in Midterm by thopd (12.1k points)
ID: 36752 - Available when: 2017-11-15 14:10 - Due to: Unlimited
| 3k views
0 like 0 dislike
69 answers
[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
| 12.9k views
0 like 0 dislike
21 answers
[Exam] asked Dec 9, 2017 in Midterm by thopd (12.1k points)
ID: 36754 - Available when: 2017-11-15 14:10 - Due to: Unlimited
| 5.1k views
0 like 0 dislike
22 answers
[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