0 like 0 dislike
5.1k 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 | 5.1k 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:172.69.59.98
©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
| 2.7k 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
| 8.8k 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
| 7.1k 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
| 3.1k 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