0 like 0 dislike
6.4k 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 | 6.4k 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:172.71.194.143
©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
| 3.3k 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
| 11.4k 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
| 9k 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
| 3.9k 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