0 like 0 dislike
5.2k 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 | 5.2k 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:172.69.130.52
©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
| 2.7k 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
| 9k 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
| 7.2k 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
| 3.1k 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