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:172.69.214.204
©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