1 like 1 dislike
10.8k 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 | 10.8k 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:162.159.115.10
©2016-2026

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
| 4.2k 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
| 17.6k 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
| 6.8k 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
| 7.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users