1 like 1 dislike
10.8k 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 | 10.8k 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:104.23.197.64
©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
| 4.2k 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
| 17.4k 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
| 6.8k 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
| 7.1k views
12,783 questions
183,442 answers
172,219 comments
4,824 users