0 like 0 dislike
5.1k 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)AC by (18k points)
ID: 18070 - Available when: Unlimited - Due to: Unlimited

reshown by | 5.1k 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:172.70.80.232
©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
| 2.7k 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
| 8.7k 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
| 7k 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
| 3.1k 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