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.)

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.

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.”

 

把題意弄明白，就知道這道題是水題了，由坐標(biāo) (0,0) 開始，以半圓為形狀每年侵蝕50 平方 miles ，問你從 (0,0) 開始到 (x,y) 結(jié)束需要多長時(shí)間，水題不需要太關(guān)注效率，所以變量定義上沒有深究，其他不解釋。

編譯器 C + + :
#include <iostream>
using namespace std;
#define PI 3.1415926
int main()
{
     int n,i=0,year;
  double x,y,area;     
     cin>>n;
     while (i<n)
  {
         cin>>x>>y;
         area = 0.5 * PI * (x*x+y*y); // semicircle area equation
         year = area/50;
         printf("Property %d: This property will begin eroding in year %d.\n",i+1,year+1);
   i++;
     }
     printf("END OF OUTPUT.\n");
  return 0;
}