计算离散化后的圆的面积损失控制在一定范围内,应该离散化为多少个边

#include "pch.h"
#include <iostream>
constexpr auto PI = (3.1415926535897932384626433832795);
using namespace std;

//计算离散化后的面积损失
double getMinusArea(double rad,int n)
{
	//圆形面积
	double cirArea = rad * rad*PI;
	//多边形面积
	double ang = PI / n;
	double rectArea = n * (rad*rad*sin(ang)*cos(ang));
	return cirArea - rectArea;
}
int main()
{
	int rad = 50;//半径
	int minArea = 5;//面积差
	
	//圆形面积
	int resNum = 0;
	int num = 1;
	while (true)
	{
		double dminArea = getMinusArea(rad, num++);
		if (dminArea < minArea)
		{
			resNum = num;
			break;
		}
	}
	cout << "n = " << resNum << endl;
}

 

 

posted on 2020-01-15 16:19  treesky  阅读(95)  评论(0编辑  收藏  举报