0 like 0 dislike
5.2k 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)AC by (18k points)
ID: 18070 - Available when: Unlimited - Due to: Unlimited

reshown by | 5.2k 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:172.70.194.139
©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
| 9k 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
| 7.2k 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
| 3.2k 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