hdu 6206 Apple（Java大数）_hdu6206a

Apple

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 806 Accepted Submission(s): 267

Problem Description

Apple is Taotao's favourite fruit. In his backyard, there are three apple trees with coordinates

(x1,y1) ,

(x2,y2) , and

(x3,y3) . Now Taotao is planning to plant a new one, but he is not willing to take these trees too close. He believes that the new apple tree should be outside the circle which the three apple trees that already exist is on. Taotao picked a potential position

(x,y) of the new tree. Could you tell him if it is outside the circle or not?

Input

The first line contains an integer

T , indicating that there are

T(T≤30) cases.
In the first line of each case, there are eight integers

x1,y1,x2,y2,x3,y3,x,y , as described above.
The absolute values of integers in input are less than or equal to

1,000,000,000,000 .
It is guaranteed that, any three of the four positions do not lie on a straight line.

Output

For each case, output "Accepted" if the position is outside the circle, or "Rejected" if the position is on or inside the circle.

Sample Input

  
   3
-2 0 0 -2 2 0 2 -2
-2 0 0 -2 2 0 0 2
-2 0 0 -2 2 0 1 1

Sample Output

  
   Accepted
Rejected
Rejected

分析：在做这种计算几何题目的时候，首先模板很重要，其次你选用的数据类型也很重要。

在这个题目中找个好几个模板，最后确定为点击打开链接；但是里面的除法对于Java大数来说一个变量是不能除的。

但是这个题目要求的是判断是否在圆的内部，所以干脆直接都乘以G再做比较，避免了精度缺失。

代码如下：

import java.math.BigDecimal;
import java.util.Scanner;


public class Main {

    public static void main(String[] args) {
        Scanner cin=new Scanner(System.in);
        BigDecimal A,B,C,G;
        BigDecimal x1,x2,x3,y1,y2,y3,x0,y0;
        BigDecimal X,Y;
        int t;
        t=cin.nextInt();
        while(t-->0)
        {
            x1=cin.nextBigDecimal();
            y1=cin.nextBigDecimal();
            x2=cin.nextBigDecimal();
            y2=cin.nextBigDecimal();
            x3=cin.nextBigDecimal();
            y3=cin.nextBigDecimal();
            x0=cin.nextBigDecimal();
            y0=cin.nextBigDecimal();
            A=x1.multiply(x1).add(y1.multiply(y1));
            B=x2.multiply(x2).add(y2.multiply(y2));
            C=x3.multiply(x3).add(y3.multiply(y3));
            G=BigDecimal.valueOf(1);
            G=y3.subtract(y2).multiply(x1).add(y1.subtract(y3).multiply(x2)).add(y2.subtract(y1).multiply(x3));
            X=B.subtract(C).multiply(y1).add(C.subtract(A).multiply(y2)).add(A.subtract(B).multiply(y3)).divide(BigDecimal.valueOf(2));
            Y=C.subtract(B).multiply(x1).add(A.subtract(C).multiply(x2)).add(B.subtract(A).multiply(x3)).divide(BigDecimal.valueOf(2));
            x0=x0.multiply(G);
            y0=y0.multiply(G);
            x1=x1.multiply(G);
            y1=y1.multiply(G);
            BigDecimal a=x0.subtract(X).multiply(x0.subtract(X)).add( y0.subtract(Y).multiply(y0.subtract(Y)));
            BigDecimal b=x1.subtract(X).multiply(x1.subtract(X)).add( y1.subtract(Y).multiply(y1.subtract(Y)));
            if(a.compareTo(b)>0)
                System.out.println("Accepted");
            else
                System.out.println("Rejected");
        }
    }
}