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:108.162.216.71
©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