0 like 0 dislike
2k 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)AD by (18k points)
ID: 18069 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00

reshown by | 2k views

13 Answers

0 like 0 dislike
Hidden content!
** * * * ** *
* ** *



int main()

{
*** * ** * **** ** x1 , y1 , x2 , y2;
**** *** * * *** a, b , ac , bc , c1 ,c2 ,c ;
** * * * ** *
** * * *** ** *** * * , &x1);

** * * * **** * ** ** *** , &y1);
**** *** ** ** * * * * * , &x2);

** **** * * **** * * * * , &y2);
****** * *** ** *
***** *** **** * ****
* ** ********** ***
** ** ** * ** ** *
* *** ** *
** * * ** * * ** **** **
** *** *** ***** *
* *** * ** ** * * *

   
** * ** ** * * * **
* **** **** *** *
* * * *** *** **
*** * *** *** ***
** *** * ** * * * ** %.6f " , a , b);
* * * * * *** ** ** *** * * ** , c);

  
*** **** ** ***
* ***** * * ** ** ** * * ** * ***** *
* ** * ** * ** ** ** * *
*** * **** ** * * 0;
* ** ** * ** ** *

}
answered by (-186 points)
0 like 0 dislike
Hidden content!
#include<stdio.h>

#include<stdlib.h>



int main ()

{
** * * *** *** ** *** * a=0,
* ** *** * * ** ** * ** * * ** ** * * *** ** * ** * *
*** *** ** ****** * * * ** ** *** ** ** ** * *
* ** * * * * **** ** ** ** * * * * * * **** ***** * **
* * *** ** ** * * *** ** **** ** * **** ** *** ** * * **
** * * * * ** * * **** ********* * * * *** ** * ****
* * **** *** *** * ** * * *** * * ** *** ***** * ** * *** *
** ***** ** **** x=0,
** ** **** ******* * * ** ** * * * ****** ****
* **** ** *** **** * *** * ** ***** * * * *
** **** * ** ** * *
** * * ***** * ** * * * ** * *** ** * * &i1, &i2, &i3, &i4);
**** *** ** * ** * *
** ** * ** ** ** ** ** = (i3 - i1);
* ** *** * *** * * = (i4 - i2);
* ** * * ** ** * *
***** ** ****** **** * ** <= x*x) && (f*f <= y*y);f++)
* * * * * ** * ** * ((x%f == 0) && (y%f == 0)){
*** * **** * **** ** * = x/f;
** *** ** ** * * * * ** = y/f;
******** ***** * * **
* ** *** * * *****
****** *** * * ** * * ** = y;
* ****** * * * * * ** = x;
** *** * ** * ***** * * = -(a*i1+b*i2);
** * ** * ** * *** **
****** * * *** * ** * *** *** *** %.6f %.6f", a, b, c);

    
* ** * *** *** * ** * * *** * * ** **
** ** * * * * * * ***** 0;

}
answered by (-170 points)
0 like 0 dislike
Hidden content!
#include * * ** * * **

#include ** *** ** * * **



int main()

{
** * *** * ** *** ** * x1 , x2 , y1 , y2 , a , b , c ;
******* *** ** ** *
** * ** * ** *** ** ** * ** ** * * * %f %f *** ***** ** ** ** * * * * *** *
** * *** *** * ** = y2 - y1 ;
* ** ** *** * *** = x1 - x2 ;
* * * * *** * ** *** * * = y1*(x2-x1)-x1*(y2-y1);
** * ** * *** ** * ** ****** * *
****** ** * ****** *
** * ** ** * *** * ******** %f %f",a,b,c);
*** ** ***** ** *** ** **** * *** *** ** * ****
**** * **** * **** ******* 0;

}
answered by (-162 points)
Welcome to Peer-Interaction Programming Learning System (PIPLS) LTLab, National DongHwa University
English 中文 Tiếng Việt
IP:172.70.179.203
©2016-2025

Related questions

0 like 0 dislike
5 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18074 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 1.3k views
0 like 0 dislike
96 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18072 - Available when: Unlimited - Due to: Unlimited
| 9k views
0 like 0 dislike
67 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18068 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 6.8k views
0 like 0 dislike
11 answers
[Exercise] Coding (C) - asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18065 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00
| 2.1k views
0 like 0 dislike
0 answers
[Resource] asked Dec 1, 2016 in 2016-1 程式設計(一)AD by Shun-Po (18k points)
ID: 18084 - Available when: Unlimited - Due to: Unlimited
| 7 views
12,783 questions
183,442 answers
172,219 comments
4,824 users