1 like 1 dislike
9.6k 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 | 9.6k 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:104.23.197.65
©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
| 3.9k 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
| 16.2k 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
| 6.4k 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
| 6.6k views
12,783 questions
183,442 answers
172,219 comments
4,824 users