0 like 0 dislike
8.2k 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 | 8.2k 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:162.159.115.10
©2016-2026

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
| 4.3k 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
| 14.3k 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
| 12k 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
| 5k 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
183,442 answers
172,219 comments
4,824 users