0 like 0 dislike
6.1k 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 | 6.1k 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:172.69.7.59
©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
| 3.2k 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
| 10.8k 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
| 8.6k 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
| 3.8k 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