1 like 1 dislike
6.8k 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 | 6.8k 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:172.69.17.67
©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
| 2.7k 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
| 11.9k 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
| 4.7k 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