0 like 0 dislike
2.1k 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)AD by (18k points)
ID: 18069 - Available when: 2016-12-01 18:30:00 - Due to: 2016-12-01 21:00:00

reshown by | 2.1k 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:172.70.80.21
©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
| 1.4k 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
| 9.6k 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
| 7.1k 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
| 2.2k 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