0 like 0 dislike
2.7k 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 | 2.7k 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:172.69.17.131
©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
| 1.4k 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
| 4.6k 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
| 3.5k 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
| 1.5k 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
183,443 answers
172,219 comments
4,824 users