Description：

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.)

Input：

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.

Output：

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.”  

Sample Input：

2

1.0 1.0

25.0 0.0

Sample Output：

Property 1: This property will begin eroding in year 1.

Property 2: This property will begin eroding in year 20.

END OF OUTPUT.

解题思路：

这道题的大致意思就是说，以原点（0，0）开始，每年50平方英里的速度开始向外扩展，问经过多少年才能达到样例输入中点的坐标的位置，那么我们直接求出样例给出的坐标点对应的半圆的面积，然后除以50就可以了（注意题中说数据的第一年从1开始），所以我们最终求出的年数要加1才是正确的！！！

程序代码：  

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
const double PI=3.1415926;
int main()
{
  int t,ans=0;
  double x,y;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%lf %lf",&x,&y);
    int year=0;
    year=(PI*(x*x+y*y)/2)/50+1;
    printf("Property %d: This property will begin eroding in year %d.\n",++ans,year);
  }
  printf("END OF OUTPUT.\n");
  return 0;
}