0 like 0 dislike
6.1k 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 | 6.1k 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:172.69.17.58
©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
| 3.2k 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
| 10.8k 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
| 8.6k 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
| 3.8k 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