1 like 1 dislike
1.2k 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 | 1.2k 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:172.71.254.145
©2016-2024

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
| 511 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
| 2k 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
| 961 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
| 970 views
12,783 questions
183,443 answers
172,219 comments
4,824 users