0 like 0 dislike
918 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 | 918 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:172.70.100.48
©2016-2024

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
| 464 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
| 1.5k 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
| 1.1k 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
| 486 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