POJ1005 I Think I Need a Houseboat【π】

题目：

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.

Hint

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.

思路：

这题很简单。意思就是：

题目：密西西比河岸某处陆地因为河水侵蚀，每年陆地面积都在减少，每年减少50平方英里，减少的陆地面积呈半圆形，即该半圆形面积以每年50平方英里的速度增长。在第一年初时，该半圆形面积为0，半圆形的圆心坐标为（0,0），现在的任务是给定一个坐标，求出该坐标在哪一年年底会被河水侵蚀。假设给定的坐标到圆心的距离不会等于半圆的半径。
输入：第一行输入坐标个数，第二行开始输入坐标，每个坐标占一行，每个坐标由横坐标和纵坐标组成，坐标值为浮点数，单位为英里。
输出：输出该坐标会在哪一年年底开始被河水侵蚀。对应每个坐标输出一行，输出结束后输出 “END OF OUTPUT.”

解题思路：
每年半圆面积增加50m²。将输入的坐标点与圆心连成的线作半径，求出这个半圆的面积，再除以50，取 比结果大的最小整数，就知道多少年后这个点被侵蚀了。

代码：

#include<iostream>
#define _USE_MATH_DEFINES
#include<math.h>
using namespace std;

int main(){
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
		double x,y;
		cin>>x>>y;
		double r=sqrt(x*x+y*y);
		double s=0.5*M_PI*r*r;
		int year=ceil(s/50);
		cout<<"Property "<<(i+1)<<": This property will begin eroding in year "<<year<<"."<<endl;
	}
	cout<<"END OF OUTPUT."<<endl;
	return 0;
}

这题在于PI在C++中如何表示。不得不说，C++在封装这块真是太不如java了。。。T_T

第一种方法，自己定义：

#define PI 3.1415926 

第二种方法（也是我在代码中使用的）：

在#include<math.h>之前，加入#define _USE_MATH_DEFINES

之后就可以使用M_PI来表示π啦~

下面我们来看看math.h中定义的关于π的宏：

//math.h
........................

#if defined(_USE_MATH_DEFINES) && !defined(_MATH_DEFINES_DEFINED)
#define _MATH_DEFINES_DEFINED

/* Define _USE_MATH_DEFINES before including math.h to expose these macro
 * definitions for common math constants.  These are placed under an #ifdef
 * since these commonly-defined names are not part of the C/C++ standards.
 */

/* Definitions of useful mathematical constants
 * M_E        - e
 * M_LOG2E    - log2(e)
 * M_LOG10E   - log10(e)
 * M_LN2      - ln(2)
 * M_LN10     - ln(10)
 * M_PI       - pi
 * M_PI_2     - pi/2
 * M_PI_4     - pi/4
 * M_1_PI     - 1/pi
 * M_2_PI     - 2/pi
 * M_2_SQRTPI - 2/sqrt(pi)
 * M_SQRT2    - sqrt(2)
 * M_SQRT1_2  - 1/sqrt(2)
 */

#define M_E        2.71828182845904523536
#define M_LOG2E    1.44269504088896340736
#define M_LOG10E   0.434294481903251827651
#define M_LN2      0.693147180559945309417
#define M_LN10     2.30258509299404568402
#define M_PI       3.14159265358979323846
#define M_PI_2     1.57079632679489661923
#define M_PI_4     0.785398163397448309616
#define M_1_PI     0.318309886183790671538
#define M_2_PI     0.636619772367581343076
#define M_2_SQRTPI 1.12837916709551257390
#define M_SQRT2    1.41421356237309504880
#define M_SQRT1_2  0.707106781186547524401

#endif  /* _USE_MATH_DEFINES */