0 like 0 dislike
6.9k 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 | 6.9k 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:104.23.197.95
©2016-2026

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
| 3.6k 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
| 12.2k 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
| 9.8k 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
| 4.2k 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