1 like 1 dislike
8k 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 | 8k 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:172.69.214.38
©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
| 3.4k 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
| 14.3k 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
| 5.6k 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
| 5.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users