0 like 0 dislike
7.2k 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 | 7.2k 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:162.159.115.9
©2016-2026

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
| 3.8k 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
| 12.8k 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
| 10.4k 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
| 4.4k 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