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:172.70.179.42
©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