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