Square Card

The first line contains a number $T(1 ≤ T ≤ 20)$ , the number of testcases.
For each testcase, there are three lines.
In the first line there are three integers $r_1,x_1,y_1(−1000 ≤ x_1, y_1 ≤ 1000, 0 < r_1 ≤ 1000)$ , the radius and the postion of the Scoring Area.
In the second line there are three integers r2,x2,y2(−1000 ≤ x2, y2 ≤ 1000, 0 < r2 ≤ 1000), the radius and the postion of the Bonus Area.
In the third line there is an integer a(0 < a ≤ 1000), the side length of the square card.
It is guaranteed that the Scoring Area is big enough thus there is always a chance for the play to be scored

For each test case, output a decimal $p(0 ≤ p ≤ 1)$ in one line, the ratio of the possibility of being scored and getting the bonus simultaneously to the possibility of being scored.

Your answer should be rounded to 6 digits after the decimal point.

C++(clang++ 11.0.1) 解法, 执行用时: 2ms, 内存消耗: 452K, 提交时间: 2022-09-23 10:21:10

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const double pi = acos(-1);
typedef long long ll;

struct point
{
    double x, y;
};

double calc(point a, double r1, point b, double r2)
{
    double dist = sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
    if (r1 + r2 <= dist) return 0;
    
    if (r1 > r2) swap(r1, r2);
    if (r2 - r1 >= dist) return pi * r1 * r1;
    
    double ang1 = acos((r1 * r1 + dist * dist - r2 * r2) / (2.0 * r1 * dist));
    double ang2 = acos((r2 * r2 + dist * dist - r1 * r1) / (2.0 * r2 * dist));
    return ang1 * r1 * r1 + ang2 * r2 * r2 - dist * r1 * sin(ang1);
}
         
int main()
{
    int T;
    scanf("%d", &T);
    
    while (T -- )
    {
        double r1, x1, y1, r2, x2, y2, aa;
        scanf("%lf%lf%lf%lf%lf%lf%lf", &r1, &x1, &y1, &r2, &x2, &y2, &aa);
        if (r1 * r1 - aa * aa / 4 <= 0 || r2 * r2 - aa * aa / 4 <= 0)
        {
            cout << "0.000000" << endl;
            continue;
        }
        
        point a = {x1, y1}, b = {x2, y2};
        double r3 = sqrt(r1 * r1 - aa * aa / 4) - aa / 2, r4 = sqrt(r2 * r2 - aa * aa / 4) - aa / 2;
        double res = calc(a, r3, b, r4);
        printf("%.6lf\n", res / (pi * r3 * r3));
    }
                 
    return 0;
}

C++(g++ 7.5.0) 解法, 执行用时: 2ms, 内存消耗: 296K, 提交时间: 2022-10-04 20:22:37

#include<bits/stdc++.h>
#define P pair<double, double>
using namespace std;
const double pi = acos(-1);
//求两圆相交面积
double area(P a, double r1, P b, double r2)
{
	double d = sqrt((a.first-b.first)*(a.first-b.first) + (a.second-b.second)*(a.second-b.second));
	if(d >= r1+r2) return 0;
	if(r1 > r2) swap(r1,r2);
	if(r2-r1>=d) return pi*r1*r1;
	double ang1 = acos((r1*r1+d*d-r2*r2)/(2*r1*d));
	double ang2 = acos((r2*r2+d*d-r1*r1)/(2*r2*d));
	return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1);
}

int main()
{
    int T;  
    cin>>T;
    while(T--)
    {
        double ra,xa,ya,rb,xb,yb,a;
		cin>>ra>>xa>>ya>>rb>>xb>>yb>>a;
		if(rb*rb-a*a/4 < 0){ cout<<"0.000000"<<endl; continue; }
        P ap=P(xa,ya), bp=P(xb,yb);
		double r1 = sqrt(ra*ra-a*a/4)-a/2, r2 = sqrt(rb*rb-a*a/4)-a/2;
		double res = area(ap, r1, bp, r2);
		printf("%.6f\n", res/(pi*r1*r1));
    }
    return 0;
}

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