1 like 1 dislike
11.9k 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 | 11.9k 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:104.23.243.108
©2016-2026

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
| 4.4k 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
| 18.5k 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
| 7.1k 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
| 7.4k views
12,783 questions
183,442 answers
172,219 comments
4,824 users