0 like 0 dislike
6.3k 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 | 6.3k 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:172.70.80.124
©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
| 11.3k 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
| 8.9k 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