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:172.69.59.218
©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