2016多校训练Contest10: 1002 Hard problem hdu5858

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#微积分

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本文介绍了一道针对小学生的奥林匹克数学题目，该题目要求计算特定几何图形中阴影部分的面积。通过给出的边长信息，利用编程方法精确计算出答案，并提供了样例输入输出以验证解决方案的有效性。

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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

似乎是从中间切开两边用积分求一下。。具体的我不太会推

另外还可以用各种软件直接弄出来

#include <stdio.h>
#include <math.h>
struct point{
    double x,y;
};
double distance(point p1,point p2){
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double area_of_overlap(point c1, double r1, point c2, double r2){
    double a = distance(c1, c2), b = r1, c = r2;
    double cta1 = acos((a * a + b * b - c * c) / 2 / (a * b)),
           cta2 = acos((a * a + c * c - b * b) / 2 / (a * c));
    double s1 = r1*r1*cta1 - r1*r1*sin(cta1)*(a * a + b * b - c * c) / 2 / (a * b);
    double s2 = r2*r2*cta2 - r2*r2*sin(cta2)*(a * a + c * c - b * b) / 2 / (a * c);
    return s1+s2;
}
int main(void){
    int t;
    int n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        point a,b;
        a.x = 0.5*n;
        a.y = 0.5*n;
        b.x = 0;
        b.y = 0;
        double ans = area_of_overlap(a,0.5*n,b,n);
        printf("%.2f\n",2*(0.5*n*0.5*n*acos(-1) - ans));
    }
    return 0;
}