0 like 0 dislike
1k views
直線方程式的通式為 ax+by+c=0 ,請設計一程式,由使用者輸入平面上兩點,程式會計算出直線方程式後輸出a, b, c。

輸入說明:輸入會包含四個浮點數,倆倆之間會用一個空白隔開,四個浮點數依序代表第一個點的x座標、第一個點的y座標、地按個點的x座標、第二個點的y座標。

輸出說明:請依序輸出 a, b, c 三個常數,每個數都需輸出到小數點後六位。請確保a, b, c 間的最大公因數為1。

輸入範例:

1.0 1.0 1.0 2.0

輸出範例:

1.000000 0.000000 -1.000000
[Exercise] Coding (C) - asked in 2016-1 程式設計(一)AC by (18k points)
ID: 18070 - Available when: Unlimited - Due to: Unlimited

reshown by | 1k views

49 Answers

0 like 0 dislike
Hidden content!
*** *** * ***


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


** * * *** *** *


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




* * ** ( * * *
*
** * **
* ******* ** ** *** * ** * *** * * ****
* = d -
= c -
** = f * b - a * e ;
* = 0 - *
* * * ***
*
* * **** ** * ** ** = - ** ** *



** *** ** * ** * * * **
**
answered by (-276 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <math.h>



int main()

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


**** * ** ** ******* * **** * (a-c!=0)
** * * * ** ****** **
** * *** ****** **** ** * ****
* *** * *** * * ** ** ** * * **
* * ****** * * ***** * * * * *
*** * * **** * * ** ****
** ** * * * ******* * ** * * ** (a-c==0)
* ** * ** **** * * ** *
**** ** * ** ** * ***** *** * * *** *
****** * *** ** * *** * ** * ***** * ** ** * *** *
* * ** * * ** * * * **** **** ***
******* * *** *** **** *
* * * * * * * * ("%.6f ",e);
* * *** * * *** *** * ("%.6f ",g);
** ***** * *** ("%.6f",f);






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

}
answered by (-136 points)
0 like 0 dislike
Hidden content!
#include *** ** * *



int main(void){


* *** *** **** ax , ay ,bx ,by ,a ,b * ** ****


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


** ** * ** * *** * - bx) != 0){
* * ** * ** * = (ay - *****
* ** * * * * * = -m*ax + ay;
** * ** ** ** ** = m;
* **** **** * = -1;

 }
** * ** ** * * *
*** *** * = 0;
** * * * * * * * *** * = 1;
* * ** ** **** = -ax;
** ** **


** * ** * * * = a;
*** * *** *** * = b;
* **** * *** * ** ** = c;


* * * * < 0)
* * * ***** **** * * * = -i;
****** * ** if (k < 0)
* ** * ***** *** ** *** = -k;


* ** ** (i < 1){
* ********* * * * = 1/i;
** * * * *** * = a*g;
* ** ** * ** ** = b*g;
* **** ***** = c*g;
* * *
** * ** * if (k < 1){
* * * ***** *** = 1/k;
* * * * * * = a*g;
* * * * ** ****** * = b*g;
* * * * **** * = c*g;
**** **
*** ** ** ** ** ** * *
* **** * ** * ****
**** * * * ** * *



return 0;

}
answered by (-122 points)
0 like 0 dislike
Hidden content!
***** ** **** *


* **** *


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


* *** * * * *




* * ( )

**** ** *
*** ****** * *** *** * * **** * * * * * * * * *
* = *
** = c - **
* = f * b - a * e ;
* = 0 -
* * * *
*
** ** *** * *** * * * = 0 - = 0 - * = *
*

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




* * ** *** ** * ** ******
* *
answered by (-276 points)
0 like 0 dislike
Hidden content!
#include ***** **





int main()

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








* ** * *** ** ** * ** ***** ** %.6f %.6f",a,b,c);

}
answered by (-162 points)
0 like 0 dislike
Hidden content!
** * ********


* **
*** * * * ** * ** * * * * * * * t = 0, a, b, c, a1 , **


*** *** ** * * *** ** *** %f %f * ** *** * * *** * **** * ** * * ** ***




***** ** ** * * **** = y2 - y1 / * * y2 - x2 * **
** ** * * * *** = * - * / * * * - * *
* * * * ** **** * **** = - * * * - ** * * *
* *** *** * * ** ** * * = a / *
* ** * *** ** = b / c;
* * * ***** *** ** ** ** *** * * * * * **


* * 0;
**
answered by (-105 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>
*** * * ******

int main (void)

{
** ** ** * ** * * ** x1,y1,x2,y2;


* ** * ** * * **** * * a,b,c;


** ** * * * ** **** * **** %f * ***** * * * * * * ** * * * ** **
** *** * * * * * * * * ****** * * * ** **
** *** ** * * ***** *** *
** * * *** *** *** **** * * ** * ** * * *




*** * * ** ** *** *** * *** * * %6f %6f",a,b,1);
* * ** **** *** ** * * 0;



}
answered by (-85 points)
0 like 0 dislike
Hidden content!
#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

#include <math.h>



int main()

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


* * *** * * *** * * (a-c!=0)
******* * ** * ** ** * * * *
* **** ** **** * * * **** **** ** **
**** *** * ** *** ****** * * **
** * ** ****** * **** * *** * ** * * ***
** * * ** ***** **** * ** *
******* * *** * *** * (a-c==0)
* * ****** ** * * **** **** *** *
* * * * * * ** *** * ** * ** ******* * * ***
* ** ****** * ** * * * **** * * ****** * * **
** ** ** ** * * * * **** ** * ** ** *** **** *
* ** * * ** **** * *
* *** * **** * ("%.6f ",e);
** * ***** ***** * * ("%.6f ",g);
* ***** ** *** ******* ("%.6f",f);






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

}
answered by (-136 points)
0 like 0 dislike
Hidden content!
* ** *** * * *** *





int * *****

{
** *** * ** ** * *
* *** * * * * %f %f * * * *** * * ** *** * * *
** ** - x2 == 0)
*** ** ******
* **** * **** *** ** = x1;
*** ** ** ** * **** ** *** ** = 0;
**** * * * *** ** * = -x1;
* * * **** * *** *** ** * *** ** * * * %f * **** **
******* * *
* ** if(x1 - x2 != 0,y1 != 0)
*** **
** ** ** * * = (y1 - ** - x2);
** *** ** * * = y1 - (a * x1);
**** ** ****** * * = -((a * x1) + c) / y1;
* * *** * ** * ****** * %f * * * * * *
* *
* * * ** 0;

}
answered by (-100 points)
0 like 0 dislike
Hidden content!
** * ******


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


** ** * **** *


* * ** * * ***




* ** ( *** )

** ** *
**** * * * ** *** ** ** * *** * ** * * **
* = d -
* = c - *
* = f * b - a * e ;


** *
*
* ******* * * * * = 0 - * = * * -
*


******* ** * * ****
** *
answered by (-276 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.69.7.14
©2016-2024

Related questions

0 like 0 dislike
16 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18075 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 508 views
0 like 0 dislike
86 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18071 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 1.7k views
0 like 0 dislike
62 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18067 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 1.2k views
0 like 0 dislike
21 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18066 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 542 views
0 like 0 dislike
0 answers
[Resource] asked Dec 1, 2016 in 2016-1 程式設計(一)AC by Shun-Po (18k points)
ID: 18085 - Available when: Unlimited - Due to: Unlimited
| 10 views
12,783 questions
183,443 answers
172,219 comments
4,824 users