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:104.23.190.141
©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