0 like 0 dislike
5.8k 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 | 5.8k 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:172.69.130.53
©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
| 3k 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
| 10k 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
| 8k 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
| 3.5k 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