0 like 0 dislike
740 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 | 740 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:172.70.126.80
©2016-2024

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
| 424 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
| 3.4k 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
| 2.5k 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
| 643 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
183,443 answers
172,219 comments
4,824 users