0 like 0 dislike
6.9k 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 | 6.9k 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:162.159.115.31
©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
| 12.2k 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
| 9.8k 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
| 4.2k 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