0 like 0 dislike
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 | 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:172.70.100.178
©2016-2024

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
| 508 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
| 1.7k 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
| 1.2k 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
| 542 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
183,443 answers
172,219 comments
4,824 users