0 like 0 dislike
6.8k 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 | 6.8k 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:104.23.197.94
©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
| 3.6k 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
| 12.1k 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
| 9.6k 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
| 4.1k 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