HDU 5858Hard problem

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原文链接：http://www.cnblogs.com/pk28/p/5784852.html

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#java

本文解析了一道针对小学奥林匹克数学竞赛的题目，通过计算正方形内阴影部分的面积来考验学生的几何知识。具体而言，该问题涉及计算两个相交圆形区域的面积，并给出了详细的算法实现。

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Hard problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 38 Accepted Submission(s): 23

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

Author

BUPT

求大圆，小圆面积的交x，2*（小圆面积-x）就是答案。

/* ***********************************************
Author        :guanjun
Created Time  :2016/8/18 12:33:04
File Name     :p1002.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const long double PI=acos(-1.0);  
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}
struct Round{
    double x,y;
    double r;
    double K(double x){
        return x*x;
    }
    double Dis(Round a,Round b){
        return sqrt(K(a.x-b.x)+K(a.y-b.y));
    }
    double Intersection_area(Round a,Round b){  
        double dis=Dis(a,b);  
        if(a.r==0||b.r==0||dis>=a.r+b.r)return 0;  
        else if(dis<=fabs(a.r-b.r))return PI*K(min(a.r,b.r));  
        else{  
            double angA=2*acos( (K(a.r)+K(dis)-K(b.r))/(2*a.r*dis) );  
            double angB=2*acos( (K(b.r)+K(dis)-K(a.r))/(2*b.r*dis) );  
            double areaA=K(a.r)*(angA-sin(angA))/2;  
            double areaB=K(b.r)*(angB-sin(angB))/2;  
            return areaA+areaB;  
        }  
    } 
}a,b;
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int n;
    cin>>n;
    double l;
    while(n--){
        scanf("%lf",&l);
        a.x=l/2.,a.y=l/2.,a.r=l/2.;
        b.x=l,b.y=0.,b.r=l;
        double ans=PI*l*l/2.-2*a.Intersection_area(a,b);
        printf("%.2f\n",ans);
    }
    return 0;
}