学步园

大意略。

思路：被告知两段圆弧在同一圆上，则圆是矩形的外接圆，求出圆心角后求出弧长即可。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;

const double eps = 1e-10;
const double PI = acos(-1.0);

struct Point
{
	double x, y;
	Point(double x = 0, double y = 0) : x(x), y(y) { }
	bool operator < (const Point& a) const
	{
		if(a.x != x) return x < a.x;
		return y < a.y;
	}
};

typedef Point Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }

Vector operator - (Point A, Point B) { return Vector(A.x-B.x, A.y-B.y); }

Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }

Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

int dcmp(double x)
{
	if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point &b)
{
	return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A, Point B, Point C) { return fabs(Cross(B-A, C-A)) / 2; }

Vector Rotate(Vector A, double rad)
{
	return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)
{
	Vector u = P-Q;
	double t = Cross(w, u) / Cross(v, w);
	return P+v*t;
}

bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)
{
	double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);
	double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);
	return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

bool OnSegment(Point p, Point a1, Point a2)
{
	return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

double PolygonArea(Point* p, int n)
{
	double area = 0;
	for(int i = 1; i < n-1; i++)
		area += Cross(p[i]-p[0], p[i+1]-p[0]);
	return area/2;
}

double PointDistanceToLine(Point P, Point A, Point B)
{
	Vector v1 = B-A, v2 = P-A;
	return fabs(Cross(v1, v2)) / Length(v1);
}

double PointDistanceToSegment(Point P, Point A, Point B)
{
	if(A == B) return Length(P-A);
	Vector v1 = B-A, v2 = P-A, v3 = P-B;
	if(dcmp(Dot(v1, v2) < 0)) return Length(v2);
	else if(dcmp(Dot(v1, v3) > 0)) return Length(v3);
	else return fabs(Cross(v1, v2)) / Length(v1);
}

int isPointInPolygon(Point p, Point *poly, int n)
{
	int wn = 0;
	for(int i = 0; i < n; i++)
	{
		const Point& p1 = poly[i], p2 = poly[(i+1)%n];
		if(p == p1 || p == p2 || OnSegment(p, p1, p2)) return -1;
		int k = dcmp(Cross(p2-p1, p-p1));
		int d1 = dcmp(p1.y - p.y);
		int d2 = dcmp(p2.y - p.y);
		if(k > 0 && d1 <= 0 && d2 > 0) wn++;
		if(k < 0 && d2 <= 0 && d1 > 0) wn--;
	}
	if(wn != 0) return 1;
	return 0;
}

int ConvexHull(Point *p, int n, Point *ch) //凸包
{
	sort(p, p+n);
	//n = unique(p, p+n) - p; //去重
	int m = 0;
	for(int i = 0; i < n; i++)
	{
		while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
		ch[m++] = p[i];
	}
	int k = m;
	for(int i = n-2; i >= 0; i--)
	{
		while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
		ch[m++] = p[i];
	}
	if(n > 1) m--;
	return m;
}

double PointToPoint(Point p1, Point p2)
{
	return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);
}

Point read_point()
{
	Point A;
	scanf("%lf%lf", &A.x, &A.y);
	return A;
}

const int maxn = 100010;
const double INF = 0x3f3f3f3f;

double A, B;

void solve()
{
	double q = 2*atan(B/A) * sqrt(A*A+B*B) / A;
	double x = 400 / (2+q);
	double Len = x, Wid = x * (B/A);
	printf("%.8lf %.8lf\n", Len, Wid);
}

int main()
{
	int T, times = 0;
	while(~scanf("%lf : %lf", &A, &B))
	{
		printf("Case %d: ", ++times);
		solve();
	}
	return 0;
}