0 like 0 dislike
3.4k 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 | 3.4k 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:172.69.59.26
©2016-2025

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
| 1.9k 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
| 5.8k 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
| 4.5k 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
| 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