0 like 0 dislike
3.2k 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 | 3.2k 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:172.71.254.194
©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
| 1.8k 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
| 5.4k 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
| 4.3k 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
| 1.9k 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