0 like 0 dislike
3.9k 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 | 3.9k 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:172.69.59.176
©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
| 2.2k 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
| 6.7k 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
| 5.3k 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
| 2.4k 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