1 like 1 dislike
1.2k 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 | 1.2k 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:172.69.58.78
©2016-2024

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
| 511 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
| 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
| 961 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
| 969 views
12,783 questions
183,443 answers
172,219 comments
4,824 users