题目1 : Farthest Point

时间限制:5000ms

单点时限:1000ms

内存限制:256MB

描述

Given a circle on a two-dimentional plane.

Output the integral point in or on the boundary of the circle which has the largest distance from the center.

输入

One line with three floats which are all accurate to three decimal places, indicating the coordinates of the center x, y and the radius r.

For 80% of the data: |x|,|y|<=1000, 1<=r<=1000

For 100% of the data: |x|,|y|<=100000, 1<=r<=100000

输出

One line with two integers separated by one space, indicating the answer.

If there are multiple answers, print the one with the largest x-coordinate.

If there are still multiple answers, print the one with the largest y-coordinate.

样例输入

1.000 1.000 5.000

样例输出

6 1

// Java版本

import java.util.Scanner;

public class Main {

/*

2

0 0

0 3

1.000 1.000 5.000

 */

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

        double x,y,r;

        x=scanner.nextDouble();

        y=scanner.nextDouble();

        r=scanner.nextDouble();

        int ll=(int) (x-r);

        int lr=(int)(x+r);

        int iya,iyb;

        double max=-1;

        double tmp;

        double r2=r*r;

        double result;

        int maxx=(int) x,maxy=(int) y;

        for(int ix=ll; ix<=lr; ++ix){

            //计算对应的iy

            tmp=Math.sqrt(r2-(ix-x)*(ix-x));

            iya=(int) (tmp+y)-1;

            iyb=(int) (y-tmp)-1;

            //System.out.println(iya+"  "+ iyb);

            result =( iyb-y)*( iyb-y)+(ix-x)*(ix-x);

            if(Double.compare(Math.sqrt(result), r)<=0&&Double.compare(result, max)>=0){ //大于等于

                max=result;

                maxx=ix;

                maxy=iyb;

            }

            iyb++;

            result =( iyb-y)*( iyb-y)+(ix-x)*(ix-x);

            if(Double.compare(Math.sqrt(result), r)<=0&&Double.compare(result, max)>=0){ //大于等于

                max=result;

                maxx=ix;

                maxy=iyb;

            }

            iyb++;

            result =( iyb-y)*( iyb-y)+(ix-x)*(ix-x);

            if(Double.compare(Math.sqrt(result), r)<=0&&Double.compare(result, max)>=0){ //大于等于

                max=result;

                maxx=ix;

                maxy=iyb;

            }

            result =( iya-y)*( iya-y)+(ix-x)*(ix-x);

            if(Double.compare(Math.sqrt(result), r)<=0&&Double.compare(result, max)>=0){ //大于等于

                max=result;

                maxx=ix;

                maxy=iya;

            }

            iya++;

            result =( iya-y)*( iya-y)+(ix-x)*(ix-x);

            if(Double.compare(Math.sqrt(result), r)<=0&&Double.compare(result, max)>=0){ //大于等于

                max=result;

                maxx=ix;

                maxy=iya;

            }

            iya++;

            result =( iya-y)*( iya-y)+(ix-x)*(ix-x);

            if(Double.compare(Math.sqrt(result), r)<=0&& Double.compare(result, max)>=0){ //大于等于

                max=result;

                maxx=ix;

                maxy=iya;

            }

        }

        System.out.println(maxx+" "+maxy);

        scanner.close();

    }

   public static void main2(String[] args) {

        Scanner scanner = new Scanner(System.in);

        double x,y,r;

        x=scanner.nextDouble();

        y=scanner.nextDouble();

        r=scanner.nextDouble();

        int ll=(int) (x-r);

        int lr=(int)(x+r);

        int ya=(int) (y+r);

        int yb=(int) (y-r);

        double max=-1;

        double tmp;

        double result;

        double r2=r*r;

        int maxx=(int) x,maxy=(int) y;

        for(int ix=ll; ix<=lr; ++ix){

            //计算对应的iy

            for( int iy=yb; iy<=ya; ++iy){

                result =( iy-y)*( iy-y)+(ix-x)*(ix-x);

                if(Double.compare(result, r2)<=0){ //如果在里面

                    if(Double.compare(result, max)>=0){

                        max=result;

                        maxx=ix;

                        maxy=iy;

                    }

                }

            }

        }

        System.out.println(maxx+" "+maxy);

        scanner.close();

    }

}

巴特西

题目1 : Farthest Point

描述

输入

输出

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