HDU 5858 Hard problem（计算几何）

Hard problem

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)

Problem Description

cjj is fun with math problem. One day he found a Olympic Mathematics problem for primary school students. It is too difficult for cjj. Can you solve it?


Give you the side length of the square L, you need to calculate the shaded area in the picture.

The full circle is the inscribed circle of the square, and the center of two quarter circle is the vertex of square, and its radius is the length of the square.

 

 

Input

The first line contains a integer T(1<=T<=10000), means the number of the test case. Each case contains one line with integer l(1<=l<=10000).

 

 

Output

For each test case, print one line, the shade area in the picture. The answer is round to two digit.

 

 

Sample Input

1

1

 

 

Sample Output

0.29

 

题意：

 求变成为length的正方形中阴影部分的面积;

 

思路：

建坐标系积分就好了;以参考资料中的D点为原点,然后得到两个圆的方程,联立求交点得到积分区间,然后积分,积分的时候借用三角函数;  http://zhidao.baidu.com/question/571519797

另一种解法也可以用以正方形的中心为原点,以对角线为水平轴,然后得到两个圆的方程,联立求交点得到积分区间,然后积分。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <queue>#include <map>#include <set>#include <stack>#include <vector>#include <list>#define LL long long#define eps 1e-8#define maxn 110#define mod 100000007#define inf 0x3f3f3f3f#define mid(a,b) ((a+b)>>1)#define IN freopen("in.txt","r",stdin);using namespace std;int main(int argc, char const *argv[]){    //IN;    int t; cin >> t;    while(t--)    {        double a; scanf("%lf", &a);        //double ans = a*a/4.0 * asin(sqrt(14.0)/4.0);        //ans -= a*a * asin(sqrt(14.0)/8.0);        //ans += a*a * sqrt(7.0) / 8.0;        //ans *= 2.0;        double ans = atan(sqrt(7.0)) / 4.0;        ans -= atan(sqrt(7.0)/5.0);        ans += sqrt(7.0) / 8.0;        ans *= a*a*2.0;        printf("%.2f\n", ans);    }    return 0;}