1 like 1 dislike
7.1k 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 | 7.1k 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:172.70.50.26
©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
| 12.7k 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
| 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
| 5.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users