1 like 1 dislike
9.3k 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 | 9.3k 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:104.23.197.95
©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
| 3.8k 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
| 15.6k 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
| 6.3k 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
| 6.5k views
12,783 questions
183,442 answers
172,219 comments
4,824 users