0 like 0 dislike
5.8k 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 | 5.8k 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:172.70.50.27
©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
| 3k 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
| 10.1k 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
| 8k 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
| 3.5k 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