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 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 | 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
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:108.162.241.32
©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
| 2.6k 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
| 11.6k 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
| 4.5k 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
| 4.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users