1 like 1 dislike
7.5k 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 | 7.5k 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:172.69.7.59
©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
| 3.1k 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
| 13.3k 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
| 5.3k 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
| 5.3k views
12,783 questions
183,442 answers
172,219 comments
4,824 users