0 like 0 dislike
4k 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 | 4k 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:108.162.216.179
©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
| 2.3k 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
| 6.8k 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
| 5.4k 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
| 2.4k 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