0 like 0 dislike
5.1k 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 | 5.1k 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:172.69.130.53
©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
| 8.7k 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
| 7k 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