0 like 0 dislike
8.1k 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 | 8.1k 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: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
| 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
| 14.2k 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
| 11.9k 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
| 4.9k 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