做网站都需要学什么,html代码是什么,在哪个网站做外贸生意好,北京网站设计制作关键词优化H - Square Card HDU - 7063
题意#xff1a;
有两个圆形区域#xff0c;一个是得分区域#xff0c;一个是获得奖金区域#xff0c;现在你有一个边长为a的正方形,当正方形在如果在某一时刻它严格在圆形范围内#xff0c;才算合法。 问把牌扔到任意的位置被得分和同时获得…H - Square Card HDU - 7063
题意
有两个圆形区域一个是得分区域一个是获得奖金区域现在你有一个边长为a的正方形,当正方形在如果在某一时刻它严格在圆形范围内才算合法。 问把牌扔到任意的位置被得分和同时获得奖金的可能性与被得分的可能性的比率是多少
题解
我第一反应是想直接rand得结果突然发现想多了其实就是求两个圆相交面积再比得分区域面积就行。但是注意题目要求牌必须完全在园内也就是圆心所在位置并不是整个圆而是比圆小一圈的情况所以需要求新的半径
newrsqrt(rr-aa/4)-a/2 求出两个新的r然后相交求圆面积做比就是结果
代码
#include bits/stdc.h
#include unordered_map
#define debug(a, b) printf(%s %d\n, a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pairint, int PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll 1e18;
const int INF_int 0x3f3f3f3f;
template typename T inline void read(T x)
{T f 1;x 0;char ch getchar();while (0 isdigit(ch)) {if (ch -)f -1;ch getchar();}while (0 ! isdigit(ch))x (x 1) (x 3) ch - 0, ch getchar();x* f;
}
template typename T inline void write(T x)
{if (x 0) {x ~(x - 1);putchar(-);}if (x 9)write(x / 10);putchar(x % 10 0);
}
void rd_test()
{
#ifdef ONLINE_JUDGE
#elsestartTime clock();freopen(in.txt, r, stdin);
#endif
}
void Time_test()
{
#ifdef ONLINE_JUDGE
#elseendTime clock();printf(\nRun Time:%lfs\n, (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
// 计算几何模板
const double eps 1e-8;
const double inf 1e20;
const double pi acos(-1.0);
const int maxp 1010;
//Compares a double to zero
int sgn(double x)
{if (fabs(x) eps)return 0;if (x 0)return -1;elsereturn 1;
}
//square of a double
inline double sqr(double x)
{return x * x;
}
/** Point* Point() - Empty constructor* Point(double _x,double _y) - constructor* input() - double input* output() - %.2f output* operator - compares x and y* operator - compares first by x, then by y* operator - - return new Point after subtracting curresponging x and y* operator ^ - cross product of 2d points* operator * - dot product* len() - gives length from origin* len2() - gives square of length from origin* distance(Point p) - gives distance from p* operator Point b - returns new Point after adding curresponging x and y* operator * double k - returns new Point after multiplieing x and y by k* operator / double k - returns new Point after divideing x and y by k* rad(Point a,Point b)- returns the angle of Point a and Point b from this Point* trunc(double r) - return Point that if truncated the distance from center to r* rotleft() - returns 90 degree ccw rotated point* rotright() - returns 90 degree cw rotated point* rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw*/
struct Point
{double x, y;Point(){}Point(double _x, double _y){x _x;y _y;}void input(){scanf(%lf%lf, x, y);}void output(){printf(%.2f %.2f\n, x, y);}bool operator(Point b) const{return sgn(x - b.x) 0 sgn(y - b.y) 0;}bool operator(Point b) const{return sgn(x - b.x) 0 ? sgn(y - b.y) 0 : x b.x;}Point operator-(const Point b) const{return Point(x - b.x, y - b.y);}//叉积double operator^(const Point b) const{return x * b.y - y * b.x;}//点积double operator*(const Point b) const{return x * b.x y * b.y;}//返回长度double len(){return hypot(x, y); //库函数}//返回长度的平方double len2(){return x * x y * y;}//返回两点的距离double distance(Point p){return hypot(x - p.x, y - p.y);}Point operator(const Point b) const{return Point(x b.x, y b.y);}Point operator*(const double k) const{return Point(x * k, y * k);}Point operator/(const double k) const{return Point(x / k, y / k);}//计算pa 和 pb 的夹角//就是求这个点看a,b 所成的夹角//测试 LightOJ1203double rad(Point a, Point b){Point p *this;return fabs(atan2(fabs((a - p) ^ (b - p)), (a - p) * (b - p)));}//化为长度为r的向量Point trunc(double r){double l len();if (!sgn(l))return *this;r/ l;return Point(x * r, y * r);}//逆时针旋转90度Point rotleft(){return Point(-y, x);}//顺时针旋转90度Point rotright(){return Point(y, -x);}//绕着p点逆时针旋转anglePoint rotate(Point p, double angle){Point v (*this) - p;double c cos(angle), s sin(angle);return Point(p.x v.x * c - v.y * s, p.y v.x * s v.y * c);}
};
/** Stores two points* Line() - Empty constructor* Line(Point _s,Point _e) - Line through _s and _e* operator - checks if two points are same* Line(Point p,double angle) - one end p , another end at angle degree* Line(double a,double b,double c) - Line of equation ax by c 0* input() - inputs s and e* adjust() - orders in such a way that s e* length() - distance of se* angle() - return 0 angle pi* relation(Point p) - 3 if point is on line* 1 if point on the left of line* 2 if point on the right of line* pointonseg(double p) - return true if point on segment* parallel(Line v) - return true if they are parallel* segcrossseg(Line v) - returns 0 if does not intersect* returns 1 if non-standard intersection* returns 2 if intersects* linecrossseg(Line v) - line and seg* linecrossline(Line v) - 0 if parallel* 1 if coincides* 2 if intersects* crosspoint(Line v) - returns intersection point* dispointtoline(Point p) - distance from point p to the line* dispointtoseg(Point p) - distance from p to the segment* dissegtoseg(Line v) - distance of two segment* lineprog(Point p) - returns projected point p on se line* symmetrypoint(Point p) - returns reflection point of p over se**/
struct Line
{Point s, e;Line(){}Line(Point _s, Point _e){s _s;e _e;}bool operator(Line v){return (s v.s) (e v.e);}//根据一个点和倾斜角angle确定直线,0anglepiLine(Point p, double angle){s p;if (sgn(angle - pi / 2) 0) {e (s Point(0, 1));}else {e (s Point(1, tan(angle)));}}//axbyc0Line(double a, double b, double c){if (sgn(a) 0) {s Point(0, -c / b);e Point(1, -c / b);}else if (sgn(b) 0) {s Point(-c / a, 0);e Point(-c / a, 1);}else {s Point(0, -c / b);e Point(1, (-c - a) / b);}}void input(){s.input();e.input();}void adjust(){if (e s)swap(s, e);}//求线段长度double length(){return s.distance(e);}//返回直线倾斜角 0anglepidouble angle(){double k atan2(e.y - s.y, e.x - s.x);if (sgn(k) 0)k pi;if (sgn(k - pi) 0)k- pi;return k;}//点和直线关系//1 在左侧//2 在右侧//3 在直线上int relation(Point p){int c sgn((p - s) ^ (e - s));if (c 0)return 1;else if (c 0)return 2;elsereturn 3;}// 点在线段上的判断bool pointonseg(Point p){return sgn((p - s) ^ (e - s)) 0 sgn((p - s) * (p - e)) 0;}//两向量平行(对应直线平行或重合)bool parallel(Line v){return sgn((e - s) ^ (v.e - v.s)) 0;}//两线段相交判断//2 规范相交//1 非规范相交//0 不相交int segcrossseg(Line v){int d1 sgn((e - s) ^ (v.s - s));int d2 sgn((e - s) ^ (v.e - s));int d3 sgn((v.e - v.s) ^ (s - v.s));int d4 sgn((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) -2 (d3 ^ d4) -2)return 2;return (d1 0 sgn((v.s - s) * (v.s - e)) 0) || (d2 0 sgn((v.e - s) * (v.e - e)) 0) || (d3 0 sgn((s - v.s) * (s - v.e)) 0)|| (d4 0 sgn((e - v.s) * (e - v.e)) 0);}//直线和线段相交判断//-*this line -v seg//2 规范相交//1 非规范相交//0 不相交int linecrossseg(Line v){int d1 sgn((e - s) ^ (v.s - s));int d2 sgn((e - s) ^ (v.e - s));if ((d1 ^ d2) -2)return 2;return (d1 0 || d2 0);}//两直线关系//0 平行//1 重合//2 相交int linecrossline(Line v){if ((*this).parallel(v))return v.relation(s) 3;return 2;}//求两直线的交点//要保证两直线不平行或重合Point crosspoint(Line v){double a1 (v.e - v.s) ^ (s - v.s);double a2 (v.e - v.s) ^ (e - v.s);return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));}//点到直线的距离double dispointtoline(Point p){return fabs((p - s) ^ (e - s)) / length();}//点到线段的距离double dispointtoseg(Point p){if (sgn((p - s) * (e - s)) 0 || sgn((p - e) * (s - e)) 0)return min(p.distance(s), p.distance(e));return dispointtoline(p);}//返回线段到线段的距离//前提是两线段不相交相交距离就是0了double dissegtoseg(Line v){return min(min(dispointtoseg(v.s), dispointtoseg(v.e)), min(v.dispointtoseg(s), v.dispointtoseg(e)));}//返回点p在直线上的投影Point lineprog(Point p){return s (((e - s) * ((e - s) * (p - s))) / ((e - s).len2()));}//返回点p关于直线的对称点Point symmetrypoint(Point p){Point q lineprog(p);return Point(2 * q.x - p.x, 2 * q.y - p.y);}
};
//圆
struct circle
{Point p; //圆心double r; //半径circle(){}circle(Point _p, double _r){p _p;r _r;}circle(double x, double y, double _r){p Point(x, y);r _r;}//三角形的外接圆//需要Point的 / rotate() 以及Line的crosspoint()//利用两条边的中垂线得到圆心//测试UVA12304circle(Point a, Point b, Point c){Line u Line((a b) / 2, ((a b) / 2) ((b - a).rotleft()));Line v Line((b c) / 2, ((b c) / 2) ((c - b).rotleft()));p u.crosspoint(v);r p.distance(a);}//三角形的内切圆//参数bool t没有作用只是为了和上面外接圆函数区别//测试UVA12304circle(Point a, Point b, Point c, bool t){Line u, v;double m atan2(b.y - a.y, b.x - a.x), n atan2(c.y - a.y, c.x - a.x);u.s a;u.e u.s Point(cos((n m) / 2), sin((n m) / 2));v.s b;m atan2(a.y - b.y, a.x - b.x), n atan2(c.y - b.y, c.x - b.x);v.e v.s Point(cos((n m) / 2), sin((n m) / 2));p u.crosspoint(v);r Line(a, b).dispointtoseg(p);}//输入void input(){p.input();scanf(%lf, r);}//输出void output(){printf(%.2lf %.2lf %.2lf\n, p.x, p.y, r);}bool operator(circle v){return (p v.p) sgn(r - v.r) 0;}bool operator(circle v) const{return ((p v.p) || ((p v.p) sgn(r - v.r) 0));}//面积double area(){return pi * r * r;}//周长double circumference(){return 2 * pi * r;}//点和圆的关系//0 圆外//1 圆上//2 圆内int relation(Point b){double dst b.distance(p);if (sgn(dst - r) 0)return 2;else if (sgn(dst - r) 0)return 1;return 0;}//线段和圆的关系//比较的是圆心到线段的距离和半径的关系int relationseg(Line v){double dst v.dispointtoseg(p);if (sgn(dst - r) 0)return 2;else if (sgn(dst - r) 0)return 1;return 0;}//直线和圆的关系//比较的是圆心到直线的距离和半径的关系int relationline(Line v){double dst v.dispointtoline(p);if (sgn(dst - r) 0)return 2;else if (sgn(dst - r) 0)return 1;return 0;}//两圆的关系//5 相离//4 外切//3 相交//2 内切//1 内含//需要Point的distance//测试UVA12304int relationcircle(circle v){double d p.distance(v.p);if (sgn(d - r - v.r) 0)return 5;if (sgn(d - r - v.r) 0)return 4;double l fabs(r - v.r);if (sgn(d - r - v.r) 0 sgn(d - l) 0)return 3;if (sgn(d - l) 0)return 2;if (sgn(d - l) 0)return 1;}//求两个圆的交点返回0表示没有交点返回1是一个交点2是两个交点//需要relationcircle//测试UVA12304int pointcrosscircle(circle v, Point p1, Point p2){int rel relationcircle(v);if (rel 1 || rel 5)return 0;double d p.distance(v.p);double l (d * d r * r - v.r * v.r) / (2 * d);double h sqrt(r * r - l * l);Point tmp p (v.p - p).trunc(l);p1 tmp ((v.p - p).rotleft().trunc(h));p2 tmp ((v.p - p).rotright().trunc(h));if (rel 2 || rel 4)return 1;return 2;}//求直线和圆的交点返回交点个数int pointcrossline(Line v, Point p1, Point p2){if (!(*this).relationline(v))return 0;Point a v.lineprog(p);double d v.dispointtoline(p);d sqrt(r * r - d * d);if (sgn(d) 0) {p1 a;p2 a;return 1;}p1 a (v.e - v.s).trunc(d);p2 a - (v.e - v.s).trunc(d);return 2;}//得到过a,b两点半径为r1的两个圆int gercircle(Point a, Point b, double r1, circle c1, circle c2){circle x(a, r1), y(b, r1);int t x.pointcrosscircle(y, c1.p, c2.p);if (!t)return 0;c1.r c2.r r;return t;}//得到与直线u相切过点q,半径为r1的圆//测试UVA12304int getcircle(Line u, Point q, double r1, circle c1, circle c2){double dis u.dispointtoline(q);if (sgn(dis - r1 * 2) 0)return 0;if (sgn(dis) 0) {c1.p q ((u.e - u.s).rotleft().trunc(r1));c2.p q ((u.e - u.s).rotright().trunc(r1));c1.r c2.r r1;return 2;}Line u1 Line((u.s (u.e - u.s).rotleft().trunc(r1)), (u.e (u.e - u.s).rotleft().trunc(r1)));Line u2 Line((u.s (u.e - u.s).rotright().trunc(r1)), (u.e (u.e - u.s).rotright().trunc(r1)));circle cc circle(q, r1);Point p1, p2;if (!cc.pointcrossline(u1, p1, p2))cc.pointcrossline(u2, p1, p2);c1 circle(p1, r1);if (p1 p2) {c2 c1;return 1;}c2 circle(p2, r1);return 2;}//同时与直线u,v相切半径为r1的圆//测试UVA12304int getcircle(Line u, Line v, double r1, circle c1, circle c2, circle c3, circle c4){if (u.parallel(v))return 0; //两直线平行Line u1 Line(u.s (u.e - u.s).rotleft().trunc(r1), u.e (u.e - u.s).rotleft().trunc(r1));Line u2 Line(u.s (u.e - u.s).rotright().trunc(r1), u.e (u.e - u.s).rotright().trunc(r1));Line v1 Line(v.s (v.e - v.s).rotleft().trunc(r1), v.e (v.e - v.s).rotleft().trunc(r1));Line v2 Line(v.s (v.e - v.s).rotright().trunc(r1), v.e (v.e - v.s).rotright().trunc(r1));c1.r c2.r c3.r c4.r r1;c1.p u1.crosspoint(v1);c2.p u1.crosspoint(v2);c3.p u2.crosspoint(v1);c4.p u2.crosspoint(v2);return 4;}//同时与不相交圆cx,cy相切半径为r1的圆//测试UVA12304int getcircle(circle cx, circle cy, double r1, circle c1, circle c2){circle x(cx.p, r1 cx.r), y(cy.p, r1 cy.r);int t x.pointcrosscircle(y, c1.p, c2.p);if (!t)return 0;c1.r c2.r r1;return t;}//过一点作圆的切线(先判断点和圆的关系)//测试UVA12304int tangentline(Point q, Line u, Line v){int x relation(q);if (x 2)return 0;if (x 1) {u Line(q, q (q - p).rotleft());v u;return 1;}double d p.distance(q);double l r * r / d;double h sqrt(r * r - l * l);u Line(q, p ((q - p).trunc(l) (q - p).rotleft().trunc(h)));v Line(q, p ((q - p).trunc(l) (q - p).rotright().trunc(h)));return 2;}//求两圆相交的面积double areacircle(circle v){int rel relationcircle(v);if (rel 4)return 0.0;if (rel 2)return min(area(), v.area());double d p.distance(v.p);double hf (r v.r d) / 2.0;double ss 2 * sqrt(hf * (hf - r) * (hf - v.r) * (hf - d));double a1 acos((r * r d * d - v.r * v.r) / (2.0 * r * d));a1 a1 * r * r;double a2 acos((v.r * v.r d * d - r * r) / (2.0 * v.r * d));a2 a2 * v.r * v.r;return a1 a2 - ss;}//求圆和三角形pab的相交面积//测试POJ3675 HDU3982 HDU2892double areatriangle(Point a, Point b){if (sgn((p - a) ^ (p - b)) 0)return 0.0;Point q[5];int len 0;q[len] a;Line l(a, b);Point p1, p2;if (pointcrossline(l, q[1], q[2]) 2) {if (sgn((a - q[1]) * (b - q[1])) 0)q[len] q[1];if (sgn((a - q[2]) * (b - q[2])) 0)q[len] q[2];}q[len] b;if (len 4 sgn((q[0] - q[1]) * (q[2] - q[1])) 0)swap(q[1], q[2]);double res 0;for (int i 0; i len - 1; i) {if (relation(q[i]) 0 || relation(q[i 1]) 0) {double arg p.rad(q[i], q[i 1]);res r * r * arg / 2.0;}else {res fabs((q[i] - p) ^ (q[i 1] - p)) / 2.0;}}return res;}
};/** n,p Line l for each side* input(int _n) - inputs _n size polygon* add(Point q) - adds a point at end of the list* getline() - populates line array* cmp - comparision in convex_hull order* norm() - sorting in convex_hull order* getconvex(polygon convex) - returns convex hull in convex* Graham(polygon convex) - returns convex hull in convex* isconvex() - checks if convex* relationpoint(Point q) - returns 3 if q is a vertex* 2 if on a side* 1 if inside* 0 if outside* convexcut(Line u,polygon po) - left side of u in po* gercircumference() - returns side length* getarea() - returns area* getdir() - returns 0 for cw, 1 for ccw* getbarycentre() - returns barycenter**/
struct polygon
{int n;Point p[maxp];Line l[maxp];void input(int _n){n _n;for (int i 0; i n; i)p[i].input();}void add(Point q){p[n] q;}void getline(){for (int i 0; i n; i) {l[i] Line(p[i], p[(i 1) % n]);}}struct cmp{Point p;cmp(const Point p0){p p0;}bool operator()(const Point aa, const Point bb){Point a aa, b bb;int d sgn((a - p) ^ (b - p));if (d 0) {return sgn(a.distance(p) - b.distance(p)) 0;}return d 0;}};//进行极角排序//首先需要找到最左下角的点//需要重载号好Point的 操作符(min函数要用) void norm(){Point mi p[0];for (int i 1; i n; i)mi min(mi, p[i]);sort(p, p n, cmp(mi));}//得到凸包//得到的凸包里面的点编号是0$\sim$n-1的//两种凸包的方法//注意如果有影响要特判下所有点共点或者共线的特殊情况//测试 LightOJ1203 LightOJ1239void getconvex(polygon convex){sort(p, p n);convex.n n;for (int i 0; i min(n, 2); i) {convex.p[i] p[i];}if (convex.n 2 (convex.p[0] convex.p[1]))convex.n--; //特判if (n 2)return;int top convex.n;top 1;for (int i 2; i n; i) {while (top sgn((convex.p[top] - p[i]) ^ (convex.p[top - 1] - p[i])) 0)top--;convex.p[top] p[i];}int temp top;convex.p[top] p[n - 2];for (int i n - 3; i 0; i--) {while (top ! temp sgn((convex.p[top] - p[i]) ^ (convex.p[top - 1] - p[i])) 0)top--;convex.p[top] p[i];}if (convex.n 2 (convex.p[0] convex.p[1]))convex.n--; //特判convex.norm(); //原来得到的是顺时针的点排序后逆时针}//得到凸包的另外一种方法//测试 LightOJ1203 LightOJ1239void Graham(polygon convex){norm();int top convex.n;top 0;if (n 1) {top 1;convex.p[0] p[0];return;}if (n 2) {top 2;convex.p[0] p[0];convex.p[1] p[1];if (convex.p[0] convex.p[1])top--;return;}convex.p[0] p[0];convex.p[1] p[1];top 2;for (int i 2; i n; i) {while (top 1 sgn((convex.p[top - 1] - convex.p[top - 2]) ^ (p[i] - convex.p[top - 2])) 0)top--;convex.p[top] p[i];}if (convex.n 2 (convex.p[0] convex.p[1]))convex.n--; //特判}//判断是不是凸的bool isconvex(){bool s[2];memset(s, false, sizeof(s));for (int i 0; i n; i) {int j (i 1) % n;int k (j 1) % n;s[sgn((p[j] - p[i]) ^ (p[k] - p[i])) 1] true;if (s[0] s[2])return false;}return true;}//判断点和任意多边形的关系// 3 点上// 2 边上// 1 内部// 0 外部int relationpoint(Point q){for (int i 0; i n; i) {if (p[i] q)return 3;}getline();for (int i 0; i n; i) {if (l[i].pointonseg(q))return 2;}int cnt 0;for (int i 0; i n; i) {int j (i 1) % n;int k sgn((q - p[j]) ^ (p[i] - p[j]));int u sgn(p[i].y - q.y);int v sgn(p[j].y - q.y);if (k 0 u 0 v 0)cnt;if (k 0 v 0 u 0)cnt--;}return cnt ! 0;}//直线u切割凸多边形左侧//注意直线方向//测试HDU3982void convexcut(Line u, polygon po){int top po.n; //注意引用top 0;for (int i 0; i n; i) {int d1 sgn((u.e - u.s) ^ (p[i] - u.s));int d2 sgn((u.e - u.s) ^ (p[(i 1) % n] - u.s));if (d1 0)po.p[top] p[i];if (d1 * d2 0)po.p[top] u.crosspoint(Line(p[i], p[(i 1) % n]));}}//得到周长//测试 LightOJ1239double getcircumference(){double sum 0;for (int i 0; i n; i) {sum p[i].distance(p[(i 1) % n]);}return sum;}//得到面积double getarea(){double sum 0;for (int i 0; i n; i) {sum (p[i] ^ p[(i 1) % n]);}return fabs(sum) / 2;}//得到方向// 1 表示逆时针0表示顺时针bool getdir(){double sum 0;for (int i 0; i n; i)sum (p[i] ^ p[(i 1) % n]);if (sgn(sum) 0)return 1;return 0;}//得到重心Point getbarycentre(){Point ret(0, 0);double area 0;for (int i 1; i n - 1; i) {double tmp (p[i] - p[0]) ^ (p[i 1] - p[0]);if (sgn(tmp) 0)continue;area tmp;ret.x (p[0].x p[i].x p[i 1].x) / 3 * tmp;ret.y (p[0].y p[i].y p[i 1].y) / 3 * tmp;}if (sgn(area))ret ret / area;return ret;}//多边形和圆交的面积//测试POJ3675 HDU3982 HDU2892double areacircle(circle c){double ans 0;for (int i 0; i n; i) {int j (i 1) % n;if (sgn((p[j] - c.p) ^ (p[i] - c.p)) 0)ans c.areatriangle(p[i], p[j]);elseans- c.areatriangle(p[i], p[j]);}return fabs(ans);}//多边形和圆关系// 2 圆完全在多边形内// 1 圆在多边形里面碰到了多边形边界// 0 其它int relationcircle(circle c){getline();int x 2;if (relationpoint(c.p) ! 1)return 0; //圆心不在内部for (int i 0; i n; i) {if (c.relationseg(l[i]) 2)return 0;if (c.relationseg(l[i]) 1)x 1;}return x;}
};
//AB X AC
double cross(Point A, Point B, Point C)
{return (B - A) ^ (C - A);
}int main()
{//rd_test();int t;read(t);while (t--) {double r1, x1, y1;double r2, x2, y2;double a;scanf(%lf%lf%lf, r1, x1, y1);scanf(%lf%lf%lf, r2, x2, y2);scanf(%lf, a);double newr1, newr2;newr1 sqrt(r1 * r1 - a * a / 4) - a / 2;newr2 sqrt(r2 * r2 - a * a / 4) - a / 2;circle A(x1, y1, newr1);circle B(x2, y2, newr2);double sum1 A.areacircle(B);double sum2 A.area();printf(%.6lf\n, sum1 / sum2);}//Time_test();
}
