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:108.162.241.184
©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