1 like 1 dislike
6.8k 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 | 6.8k 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:172.70.80.233
©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
| 2.7k 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
| 12k 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
| 4.8k 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
| 4.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users