1 like 1 dislike
4.9k 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 | 4.9k 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:172.70.130.161
©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
| 2k 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
| 8.8k 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
| 3.4k 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
| 3.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users