0 like 0 dislike
7.5k 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 | 7.5k 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:104.23.243.108
©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
| 3.9k 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
| 13.3k 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
| 10.9k 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
| 4.6k 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