2 like 0 dislike
12.4k views

Write a program that asks the user to enter a fraction, then reduces the fraction to lowest terms.

寫一個程式 輸入一個分數,約分為最簡分數後輸出

Example input:

6/12

Example output:

1/2

Example input 2:

12/6

Example output:

2
[Exercise] Coding (C) - asked in Chapter 6: Loops by (5.2k points)
ID: 31048 - Available when: 2017-11-09 18:00 - Due to: Unlimited

edited by | 12.4k views

67 Answers

0 like 0 dislike
Hidden content!
#include <stdio.h>

int main()
{

*** * **** **** * * *** i;
*** ** *** * * ***** ** j;
***** * **** * * **** ** *** * *** newNum1;
** *** ** **** * * * newNum2;
* ** **** ** **** * ** gcd;
***** * * ** * ** * * rem;
*** * * ** *** * * num1, num2;

* * * * ** *** *** ** * * ** *** * *

** * * * ******* ** **** *
* * **** ** ***** *

* **** *** * ********* * * != 0)
* ** **** * * *** * ***
*** * * ** * * ** * * *** * * ***** ** * ** **** = j%i;
*** *** * ****** * * ** * * * * **** = i;
** * * ** * * ** **** * ****** **** = rem;
*** ** ** *** **** *** *
*** * ** * ****** ***

* * * * ** * * ** ** * * = num1/gcd;
* * ******* * ** * = num2/gcd;

* * * ******* * * * * == 1)
    {
** ** * *** * ** *** ** *** * * *** * ** * * *** ****
    }

* ** * * ***** **** * ** **
    {
***** * * * * * *** ** * * * * * * ** * **** ** **** ** *
*** * * *** * *** *

** **** ** ** **** * ** * * 0;
}
answered by (-140 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 <stdlib.h>
#include <stdio.h>

int main()
{
    int num,deno,i;
* * * * ** * * *** * *** * ** * * * ** ** *

** *** ** * * * ***** ** *** *
* ** * *** **** *** *** *** * * ****** *** ** ***** * ** **** ** ** *

    else{
***** * *** * *** * ** * **** *
    {
* ** ** *** ** ** ** * * *** ** ****
** * * ** * ******* ** *** ** * *** ** * *** ****** *** ** * * * **** && deno%i==0)
**** * * * * ** * ** * * * *** * ** ** ** ** * * ** **** *
* ** * * * ** ** * ** ** ** * * ** * ** ******** ** ***** * ** * * * * ** ** *
** * ** *** * * ** ********* ****** **** *** * * * * * *** *** ** * * ****
* * ******* * * * * ** ** **** ** **** * ** * **** * * * *** *
    }

** * * ** * ** *** * ** * *
    {
* ** ** * * ** * ** ** ** ** *
* * * * ** *** * * ** ** * * **** * * * ** ***** * * * * * && deno%i==0)
* * ****** * ** * * * * *** ***** **** ** * ** * * ** ** *** *
* * ** * * * * * ** * * *** * * * * ** * ** **** ** * **** * * * * * * **
* **** * * * * ** * *** ** *** ** * **** * ******* ** ** * *** * ** ** ** *** * **
* * * *** **** ** * *** * * * ** **** * * * * ** *
    }
* ** * ** ** * ** * * * * * * ** ** *** * * *

    }




* * ** * * ** * ** **** 0;

}
answered by (-16 points)
edited by
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
Case 4: Correct 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 * * *** * * *
#include * * ****** * ***

int gcd(int n, int m)
{
** ** ** ** * * * *** * gcd, remainder;

* *** * ** ** *** ** *** ** (n != 0)
**** * ** * * * * *
* ***** * *** ** ** ** **** ** = m % n;
* *** * * * ** * ** *** *** * ** ** **** * = n;
** ** ** * * ** * ***** * * * ** * * * * * * = remainder;
* * * * * **** * *

* * * * ** ** ** * * = m;

* **** * *** * * ** * * gcd;
}

int main (int argc, const char * argv[]) {

*** * * *** * ** *** number1, number2;
* *** * ** ** * **** * * newNumber1, newNumber2;

****** ** * ** *** ** * * ****** ** * *** *** * ****

* * * * ** ** ** ** = number1 / * * **** number2);
** ** ****** ******** * * * = number2 / * number2);

* * ** * * * * *** * **** ** ** * ***
*** *** ** * ** **** * * * ** ** ** ** * * ** * ** ****
* *** * * **** ** ** **
* * * ** * ** ** **** **** ** *** *** * *
}
answered by (-108 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!
** * ** *
** *** ** ** ** *

int ** * n, int m)
{
**** * *** * int gcd, *** *

** * * (n != 0)
** * * {
* * * * * * *** *** * * * = m % n;
* * *** * * ** * * m = n;
** ** ** * * ** ** * *** n = * **
** ** }

* * ****** gcd = m;

* ****** * *** gcd;
}

int main (int *** * char * *** {

*** * * * int ***** * ***
****** ** ** * int * *

* * * * * * * ** ** **** ** ********* * *** **** *

**** * * *** *** ** = **** / * * ** * *
* * * * *** ** = * / * * **

* * *** * ** if * *** * * * ** *
* * * * * ***** *** * **** * * * * *
* **** * * *** else
* ** *** * ************* * ** ******* ** **** **
}
answered by (-168 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()
{
 
* * * ** * ******** i;
**** * ** ** * * j;
*** * ** ** **** * * * * newNum1;
* ** ***** * ***** ** ** * * newNum2;
* * * ** *** gcd;
* * ** * ** *** * * rem;
* * * ** ** * *** * * num1, num2;
    
**** ** *** * ***** **** ** * * ***** *** **** * * ** * * *
* * * **** ***** * **** ***
*** ** **** ** * * * * ***
** *** * **** **** ** **
* ** * ** * * * *
** * * ** * * * * * *** **** != 0)
** * ** *** **** ****
* * **** **** * * * *** ** * * * * * ** ** = j%i;
* ** ** ** ***** *** * * * ** * ** * *** * = i;
** * **** * **** * **** ***** *** *** **** ** ** * = rem;
** * ***** ** *** *** ****
* * ** ** * * **** ** ** **
    
* * ***** ** * * * = num1/gcd;
* * *** ** ** * *** = num2/gcd;
    
***** ** ** **** * * ** * == 1)
*** ** * ** ** * * ****
*** * ***** * ** * * ** ** * * * ** * ** *** *** ** * * * ** *
* **** * ** ** *** *
    
* * * * * ** * *** * **
* * * *** *
** * *** * * ** ** * * ** * * * ** ******** * **
** * ** ********* * * * ** *
    
* ****** * * * * ** * * * 0;
}
answered by (-284 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()
{
 
* * * * * ** *** * * * i;
* *** * ** **** * * ** ** j;
* * ** * * * *** ** * newNum1;
* * ** ** * * * * * * * newNum2;
** ** ** *** * * ** *** ** * gcd;
* *** * * * * ***** * ** rem;
** ** ***** ** ***** num1, num2;
    
** ** * ** ** **** * ** * * ** * * * ** *** **
* ***** * ** * **** * *
* ** * * * * ***
* ** ** * ** * *** * * ******
** * * ** ** * * ****
* * * * * ** ** * * != 0)
* ** * * *** ** * *
** ***** * * *** ** * *** * *** * * * ******* = j%i;
** * *** ** * ** * * * ** ** **** *** = i;
** *** * * * *** **** ** * ** * = rem;
** * * * * * * * ***
**** ** ** * * * * * ***
    
***** ** ** **** ** * * ** = num1/gcd;
*** * ** * ** * * * = num2/gcd;
    
* **** ** *** ** * ** * * * * ** == 1)
** ** * * *** * ** *****
** * ** * * * * ****** * * *** ***** * ** ** * * ** *** * * *
** * *** * * *** **** *
    
* * ** * * * **** **** * *
* ** * * ** ** ** *
** *** ** * * ** * ** * * ** **** ** ** * ***** ** * ********* *
* * ** * * * * ***** *** * *
    
**** ** * * **** **** * *** 0;
}
answered by (-193 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 n,m,i,gcd;
** ***** ** * * * * ** ** * ** * * * *
**** * ** * * * * *** * * * **** ****** * * * * * *** *
    {
* * ** ** ** * * * ** * * ** ** ** * ** ** * * *
* *** * * * *** *** **** * * *** *** * ***
* ** ** * * * * * * * * ***** * * **
** * * * * *** * *** * **** * ** * * ***
* **** ** *** * ** *** * **

* * * * * ***** ** *** * ****
* *** *** * * * * ** *** * ***** * * * *** * ** * * ** ** ****** ** * **** ** **
* ** **** ** * * *** * **** ******* ** ** * *** *** * *** * ** *


return 0;
}
answered by (-32 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>
#include <stdlib.h>


int gcd(int,int);
int main(){
* *** **** ** ** * * *** * * a,b,GCD;
** ** ** ** *** * * ** *** * * * ** * ** * ** *
* * * * **** * **** **** *
***** ** ** **** *** * **
* * *** ***** * * ***** ** **** ** * **** ** **
* * * ****** *** *** ** * *
**** * ** * * ** ******* * * * ** ** * ** **** * * **** **
** * *** * ***** ** * ** 0;
}
* * * * ** *** * * **** ** gcd(int a, int b){
* ** **** * * * * * * * ***** ** * *** * ** * (!b) return (a);
* * * *** **** * ** ***** * * * * * * * gcd(b, a % b);
** *** * * ** * * ****
answered by (-32 points)
edited by
0 0
prog.c: In function 'main':
prog.c:6:1: warning: implicit declaration of function 'printf' [-Wimplicit-function-declaration]
 printf("\xe8\xab\x8b\xe8\xbc\xb8\xe5\x85\xa51\xe5\x80\x8b\xe5\x88\x86\xe6\x95\xb8\xef\xbc\x8c\xe5\xa6\x82:4/8\n");
 ^~~~~~
prog.c:6:1: warning: incompatible implicit declaration of built-in function 'printf'
prog.c:6:1: note: include '<stdio.h>' or provide a declaration of 'printf'
prog.c:7:1: warning: implicit declaration of function 'scanf' [-Wimplicit-function-declaration]
 scanf("%d/%d",&a,&b);
 ^~~~~
prog.c:7:1: warning: incompatible implicit declaration of built-in function 'scanf'
prog.c:7:1: note: include '<stdio.h>' or provide a declaration of 'scanf'
0 0
prog.c: In function 'main':
prog.c:7:1: warning: implicit declaration of function 'scanf' [-Wimplicit-function-declaration]
 scanf("%d/%d",&a,&b);
 ^~~~~
prog.c:7:1: warning: incompatible implicit declaration of built-in function 'scanf'
prog.c:7:1: note: include '<stdio.h>' or provide a declaration of 'scanf'
prog.c:9:1: warning: implicit declaration of function 'printf' [-Wimplicit-function-declaration]
 printf("%d/%d",a/GCD,b/GCD);
 ^~~~~~
prog.c:9:1: warning: incompatible implicit declaration of built-in function 'printf'
prog.c:9:1: note: include '<stdio.h>' or provide a declaration of 'printf'
0 0
prog.c: In function 'main':
prog.c:7:1: warning: implicit declaration of function 'scanf' [-Wimplicit-function-declaration]
 scanf("%d/%d",&a,&b);
 ^~~~~
prog.c:7:1: warning: incompatible implicit declaration of built-in function 'scanf'
prog.c:7:1: note: include '<stdio.h>' or provide a declaration of 'scanf'
prog.c:9:1: warning: implicit declaration of function 'printf' [-Wimplicit-function-declaration]
 printf("%d/%d",a/GCD,b/GCD);
 ^~~~~~
prog.c:9:1: warning: incompatible implicit declaration of built-in function 'printf'
prog.c:9:1: note: include '<stdio.h>' or provide a declaration of 'printf'
0 0
Case 0: Correct output
Case 1: Wrong output
Case 2: Correct output
Case 3: Correct 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 ** * ** ** ** **
#include ** * ***

//gcf function - return gcd of two numbers
int gcd(int n, int m)
{
**** * **** ** *** * * gcd, remainder;
** *** * * **** * * *
*** * *** ** ** * ** (n != 0)
** * * ***** * ******** **
** * ** * ** * * ** * *** * ***** ** * *** = m % n;
* * * * * ** * ** * * * * * * * = n;
**** * * ** *** ** * ** * *** * ***** * = remainder;
** *** *** * * *** * ** *
*** * ** ** *** ***
** ** * ***** * * ** * = m;
* * * * ** ** * *
** *** ** ** **** gcd;
}
int main (int argc, const char * argv[]) {
****** * * * *** * number1, number2;
** * * * * * ***** * * * newNumber1, newNumber2;
* **** * * * *
* ***** * * * ** * ** *
* * ** ** **** * ** ** ** ******* * ****** *** * ** * ** * *** **
** * ***** * * * * * *
**** ** ******** * * * ***** = number1 / gcd(number1, number2);
* * *** * ** * *** *** * * = number2 / ** ** * number2);
** * * *** * * * *** *
* ** * *** ***** ***
****** ***** * ** * *** ( * ** ** *
* *** * * **** *** * *** ******* *** ***** * ** * ** *** *** * * ** newNumber1);
** ** **** ** * ** *
* * * * * * *******
***** * ** *** ** * ****** ** * * ** * newNumber1, newNumber2);
}
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 <stdio.h>
#include <stdlib.h>

int main()
{
*** * ** *** * * * *** ** a,b,c,d,e,f;
* * * * * **** ** * ** * ****** **** * *** ** *****
**** *** *** ** ** * *
*** * * ** * * * * *
* * ** * * * ** * * * *********
    {
* * * ** * * ** * * **** * * *** ** *
*** * ****** * ******* ******* * * ** * * * **
**** ** ** * ** * *** * ** * * *
    }
** ** * **** ** * * * *
    {
* * *** ** * * ** * * * * ** * * *** **
*** * * * ** * * * *** *** * *** *** * * * * ****** * ** *
** ** ** * * *** * * ** ***** ** ** ** * **
** * ** * ** *** ** ****** * * * ** ****** ** * *** ** ****** * *
    }
**** ** ** * * ** * ** * ****
    {
*** * * * **** ***** ** * ** ** * **** * * **
**** * ** * * *** * * * *** * * ** ** * * * ***
** * * ** ******** * * * * * * * ** * *
** *** ** **** * * * ** * * ** *** * ****** * ** *** *** * * **
    }
** ** ******* ** *** * ** 0;
}
answered by (-498 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:172.69.130.13
©2016-2025

Related questions

0 like 0 dislike
71 answers
[Exercise] Coding (C) - asked Nov 9, 2017 in Chapter 6: Loops by semicolon (5.2k points)
ID: 31047 - Available when: 2017-11-09 18:00 - Due to: Unlimited
| 12.4k views
0 like 0 dislike
72 answers
[Exercise] Coding (C) - asked Nov 2, 2017 in Chapter 6: Loops by semicolon (5.2k points)
ID: 29831 - Available when: 2017-11-02 18:00 - Due to: Unlimited
| 12.3k views
1 like 1 dislike
85 answers
[Exercise] Coding (C) - asked Oct 26, 2017 in Chapter 6: Loops by semicolon (5.2k points)
ID: 28914 - Available when: 2017-10-26 18:00 - Due to: Unlimited
| 15.7k 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
| 11.9k views
12,783 questions
183,442 answers
172,219 comments
4,824 users