1 like 1 dislike
8.2k 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 | 8.2k 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:172.70.80.117
©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
| 3.5k 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
| 14.6k 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
| 5.8k 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
| 5.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users