1 like 1 dislike
6.7k 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 | 6.7k 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:172.71.254.192
©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
| 11.8k 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
| 4.6k 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
| 4.6k views
12,783 questions
183,442 answers
172,219 comments
4,824 users