0 like 0 dislike
3.9k 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 | 3.9k 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:172.70.130.204
©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
| 2.2k 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
| 6.7k 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
| 5.3k 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
| 2.4k 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