1 like 1 dislike
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 | 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:172.70.80.196
©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
| 3.4k 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
| 14.3k 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
| 5.6k 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
| 5.7k views
12,783 questions
183,442 answers
172,219 comments
4,824 users