E . Rain Gauge -UCF Local Programming Contest 2015

https://nanti.jisuanke.com/t/43390

题意

　　几何中心重合的 圆 和 正方形

　　给定正四边形边长 a 和 圆的半径 r，求重合面积

　　简单计算几何问题

思路

　　分块计算

　　注意精度和舍入问题

代码

　　

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;


bool compare(int a, int b)
{
	return a < b;//升序
}


double pi = 3.14159265358979;
int main(){
	double s = 0;
	double r = 0;
	int t = 0;
	cin >> t;
	while(t--)
	{
		cin >> s >> r;
		if(r > s*sqrt(2)/2)
		{
			//cout<<" !1 ";
			cout <<s*s <<endl;
			continue;
		}
		if(r < s/2)
		{
			//cout<<" !2 ";
			cout <<pi*r*r <<endl;
			continue;
		}
		double d = sqrt(r*r - s*s/4);
		double s1 = d*s*2;
		double a = acos(s/r/2);
		//cout<<a<<endl;////
		double s2 = pi*r*r * (1-a/(2*pi)*8); 
		cout <<fixed <<setprecision(2) <<s1 + s2 <<endl;
	}
    return 0;
}