1 like 1 dislike
7.4k 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 | 7.4k 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:172.69.130.71
©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
| 13k 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
| 5.2k 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