1005

 

总时间限制:

1000ms

内存限制:

65536kB

描述

Fred Mapper is considering purchasing some land in Louisiana to build his house on. In the process of investigating the land, he learned that the state of Louisiana is actually shrinking by 50 square miles each year, due to erosion caused by the Mississippi River. Since Fred is hoping to live in this house the rest of his life, he needs to know if his land is going to be lost to erosion.

After doing more research, Fred has learned that the land that is being lost forms a semicircle. This semicircle is part of a circle centered at (0,0), with the line that bisects the circle being the X axis. Locations below the X axis are in the water. The semicircle has an area of 0 at the beginning of year 1. (Semicircle illustrated in the Figure.)

输入

The first line of input will be a positive integer indicating how many data sets will be included (N). Each of the next N lines will contain the X and Y Cartesian coordinates of the land Fred is considering. These will be floating point numbers measured in miles. The Y coordinate will be non-negative. (0,0) will not be given.

输出

For each data set, a single line of output should appear. This line should take the form of: “Property N: This property will begin eroding in year Z.” Where N is the data set (counting from 1), and Z is the first year (start from 1) this property will be within the semicircle AT THE END OF YEAR Z. Z must be an integer. After the last data set, this should print out “END OF OUTPUT.”

样例输入

2
1.0 1.0
25.0 0.0

样例输出

Property 1: This property will begin eroding in year 1.
Property 2: This property will begin eroding in year 20.
END OF OUTPUT.

提示

1.No property will appear exactly on the semicircle boundary: it will either be inside or outside.
2.This problem will be judged automatically. Your answer must match exactly, including the capitalization, punctuation, and white-space. This includes the periods at the ends of the lines.
3.All locations are given in miles.

 

这篇文章翻译出来:

 

 

翻译有错，实属巧合！

 

描述：
Fred Mapper想要买Louisiana 的一些土地去建造他的房子，再调查土地的过程中，
他认识到Louisiana 现在正在每年以50平方米的范围缩减由于密西西比河的吞噬，
但是Fred想在靠这座房子度过他的余生，他需要知道他的土地是否会被淹没。
再做了更多的研究后，Fred认识到这里的土地正在像一个半圆形减少，这个圆形
是以（0,0）为中心，x轴把这个圆一分为2，x轴以下的都是水，这个半圆第一年的面积为0.
（半圆在下面有图解）
输入：
第一行输入一个正整数表示下面将会包含多少组数据，接下来的每一行将会包含笛卡尔坐标系
中x,y两个坐标点.这些浮点型数据都是以米(m)为单位，y坐标是非负的，不能输入(0,0).
输出：
输出每一行数据，每一行数据应该有如下格式:"Property N: This property will begin eroding in year Z.”.
 N是数据集，从1开始，Z是第一年(从1开始)，将会在结束的那一年之内。Z必须是一个整数，最后一行数据，要输出
“END OF OUTPUT.”
提示：
1.半圆的边界没有数，要么在里面，要么在外面。
2.这个问题将会被自动判断，你的答案要匹配正确，包括大写，标点，空格，每一行后面都有句号
3.所有坐标都是以m为单位给出。

 

解决方案一:G++

 

 

#include<iostream>
#include<math.h>
#include<cstdio>
#define PI 3.1415926f
using namespace std ;
int main()
{
    int n ;
    cin>>n ;
    int j=1;
    while(n--)
    {
        float x ;
        cin>>x ;
        float y ;
        cin>>y ;
        double len = sqrt(x*x+y*y) ;
        double sum = 100 ;
        int i ;
        double r = sqrt(sum/PI) ;
        for(i=0;r<len;i++)
        {
            r = sqrt(sum/PI) ;
            sum = sum+100;

        }
        if(i==0){
         i= 1 ;
        }
        printf("Property %d: This property will begin eroding in year %d.\n"
                  ,j,i) ;
        j++ ;
    }
    cout<<"END OF OUTPUT."<<endl ;
       return 0 ;

}

 

 

 

解决方案二:Java

 

import java.util.Scanner;

public class Main{

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
     Scanner sc = new Scanner(System.in) ;
     int n = sc.nextInt() ;
     int m = n;
     float x,y;
     while(m>0)
     {
    	 x = sc.nextFloat() ;
    	 y = sc.nextFloat() ;
    	 float sum = 100 ;
    	 double len = Math.sqrt((double)(x*x+y*y));
    	 double r = Math.sqrt(100/Math.PI);
    	 int i ;
       	 for(i=0; len>r;i++)
    	 {    
    		 r = Math.sqrt(sum/Math.PI);
    		 sum += 100; 
         }
       	 if(i==0){
       		 i=1 ;
       	 }
    	 System.out.println("Property "+(n-m+1)+": This property will begin eroding in year "
        		 +i+".");
    	 
    	 
    	 m-- ;
     }
     System.out.println("END OF OUTPUT.") ;
	}

}

 

 核心思想：

 

 

 

其实就是随便选一个点，求多少年这个

 

点会被淹没。