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:172.70.111.109
©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