Circle in Square LightOJ - 1022

最新推荐文章于 2022-01-29 15:16:47 发布

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本文介绍了一道关于计算圆置于正方形中时周围阴影部分面积的编程题。通过给定圆的半径，利用数学公式计算出蓝色阴影部分的面积，并提供了一个简单的C++代码实现。

题目

Circle in Square

LightOJ - 1022

A circle is placed perfectly into a square. The term perfectly placed means that each side of the square is touched by the circle, but the circle doesn't have any overlapping part with the square. See the picture below.

Now you are given the radius of the circle. You have to find the area of the shaded region (blue part). Assume that pi = 2 * acos (0.0) (acos means cos inverse).

Input

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case contains a floating point number r (0 < r ≤ 1000) denoting the radius of the circle. And you can assume that r contains at most four digits after the decimal point.

Output

For each case, print the case number and the shaded area rounded to two places after the decimal point.

Sample Input

30.091

87.0921

Sample Output

Case 1: 343.36

Case 2: 777.26

Case 3: 6511.05

Hint

This problem doesn't have special judge. So, be careful about precision problems. Better to add a small value to your result to avoid precision problems. For example, add 10^-9 to your result.

思路

水题。。。收获就是10^-9的浮点数要用double， float装不下。。。之前一直不理解题目中的意思。。。。。。

代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#include <set>
#include <vector>

#define MAX_N 0x3F3F3F3F
const double pi = 2*acos(0.0);
using namespace std;

int main()
{
	int t;
	double r;
	double res;

	cin >> t;
	for (int i = 1; i <= t; i++)
	{
		cin >> r;
		cout << "Case " << i << ": ";

		res = (4 - pi)*r*r;

		cout.setf(ios_base::fixed);
		cout.precision(2);
		cout << res << endl;
	}
	return 0;
}