1 like 1 dislike
12.3k 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 | 12.3k 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:104.23.197.65
©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
| 4.5k 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
| 19k 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
| 7.3k 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
| 7.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users