1 like 1 dislike
2.7k 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 | 2.7k 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:172.69.59.44
©2016-2024

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
| 1.2k 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
| 5k 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
| 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
| 2.1k views
12,783 questions
183,443 answers
172,219 comments
4,824 users