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:172.70.80.170
©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
| 11.7k 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
| 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