2021CCPC 桂林 F. Illuminations II （计算几何）
2024-05-10 06:18:05
F. Illuminations II
Solution
内多边形每条线与外多边形做交点，利用前缀和计算即可。
还是挺容易想到的，比赛的时候没时间写了，写计算几何再也不用cin了。
Code
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#include <bits/stdc++.h>using namespace std;
using ll = long long;#define rep(i, a, b, stp) for(int i = a; i <= b; i += stp)
#define all(a) a.begin(), a.end()namespace BASIC_MATH {// const ll mod = 998244353, mod_g = 3, img = 86583718;const ll mod = 19260817;const double eps = 1e-6; // when use double gcd, let eps smaller(1e-4)const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const int maxp = 100010;
// Very Basicll mul(ll a, ll b) { ll z = (long double) a / mod * b; ll res = (unsigned long long) a * b - (unsigned long long) z * mod; return (res + mod) % mod; }// O(1) quick_mul, use long doubleinline ll quick_pow(ll ans, ll p, ll res = 1) {for(; p; p >>= 1, ans = mul(ans, ans) % mod)if(p & 1) res = mul(res, ans) % mod;return res % mod;}double gcd(double a,double b) {if(fabs(b) < eps) return a;if(fabs(a) < eps) return b;return gcd(b, fmod(a,b));}int gcd(int a, int b) { return __gcd(a, b); }ll gcd(ll a, ll b) { return __gcd(a, b); }
// Exgcdll exgcd(ll A, ll B, ll &x, ll &y) { // Ax + By = Cif(B == 0) {x = 1; y = 0; return A; }ll _gcd_ = exgcd(B, A % B, x, y);//逆推回去求解ll t = x;x = y; y = t - A / B * y;return _gcd_;}void call_exgcd(ll A, ll B, ll C, ll &x, ll &y) {ll d = exgcd(A, B, x, y);if(C % d != 0) cout << -1 << '\n';else { // k = 0 的通解x = C / d * x;y = C / d * y;x %= mod; while(x < 0) x += mod;y %= mod; while(y < 0) y += mod;cout << x << ' ' << y << '\n';}}
// Geometryint sgn(double x) { if(fabs(x) < eps) return 0; return x > 0? 1: -1; }inline double sqr(double x) { return x * x; }struct Point {double x, y;Point() { }Point(double _x,double _y): x(_x), y(_y) { }void input() { cin >> x >> y; }void output() { cout << fixed << setprecision(12) << x << ' ' << y << '\n'; }bool operator == (Point b)const { return sgn(x - b.x) == 0 && sgn(y - b.y) == 0; }bool operator < (Point b)const { return 1ll * x * b.y < y * b.x; }Point operator -(const Point &b)const { return Point(x - b.x, y - b.y); }double operator ^(const Point &b)const { return x * b.y - y * b.x; } // 叉乘double operator *(const Point &b)const { return x * b.x + y * b.y; } // 点乘double len() { return hypot(x, y); }double len2() { return x * x + y * y; }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`double 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angle`Point 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);}};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, 0 <= angle < pi`Line(Point p,double angle) {s = p;if(sgn(angle-pi/2) == 0) e = (s + Point(0, 1));else e = (s + Point(1, tan(angle)));}//ax + by + c = 0Line(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); }double length() { return s.distance(e); }//`倾斜角 0 <= angle < pi`double 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;else return 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; }//`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:`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``相交距离为0`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); }};struct circle {Point p; double r;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``利用两条边的中垂线得到圆心`circle(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`circle(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);}void input() { p.input(); cin >> r; }void output() { cout << fixed << setprecision(9) << "x:" << p.x << ", y:" << p.y << ", r:" << r << '\n'; }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)); }double area() { return pi * r * r; }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`int 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`int 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 = r1;return t;}//`得到与直线u相切，过点q,半径为r1的圆`int 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`int 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`int 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)`int 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;}//`求两圆相交的面积(精度更高)(需要long double)`double areacircle2(circle v) {double a=hypot(p.x-v.p.x,p.y-v.p.y),b=r,c=v.r;double s1=pi*r*r,s2=pi*v.r*v.r;if(sgn(a-b-c)>=0)return 0;if(sgn(a+min(b,c)-max(b,c))<=0)return min(s1,s2);else {double cta1=2*acos((a*a+b*b-c*c)/(2*a*b));double cta2=2*acos((a*a+c*c-b*b)/(2*a*c));return cta1/(2*pi)*s1-0.5*sin(cta1)*b*b+cta2/(2*pi)*s2-0.5*sin(cta2)*c*c;}}//`求圆和三角形pab的相交面积`double 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;}};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`void norm() { Point mi = p[0]; for(int i = 1; i < n; i++)mi = min(mi,p[i]); sort(p,p+n,cmp(mi)); }//`得到的凸包里面的点编号是0~n-1``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`void 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--; 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`void 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[3];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：HDU3982`void 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 LightOJ1239`double 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`double 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]);else ans -= 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;}};
}
using namespace BASIC_MATH;void solve() {int n, m; scanf("%d%d", &n, &m);vector<Point> outer(n), inner(m);rep(i, 0, n - 1, 1) scanf("%lf%lf", &outer[i].x, &outer[i].y);rep(i, 0, m - 1, 1) scanf("%lf%lf", &inner[i].x, &inner[i].y);vector<Line> outer_line(n), inner_line(m);vector<double> siz(n);rep(i, 0, n - 1, 1) {outer_line[i] = {outer[i], outer[(i + 1) % n]};siz[i] += (i? siz[i - 1]: 0) + outer_line[i].length();}rep(i, 0, m - 1, 1) inner_line[i] = {inner[i], inner[(i + 1) % m]};int f = 0, t = 0;double ans = 0;Point temp;rep(i, 0, m - 1, 1) {Point P(inner[i]), Q(inner[(i + 1) % m]);Point cro1, cro2; Line l(P, Q);while(1) {if(l.linecrossseg(outer_line[f]) == 0) {f = (f + 1) % n;continue;}temp = l.crosspoint(outer_line[f]);if(sgn(P.distance(temp) - Q.distance(temp)) < 0) {f = (f + 1) % n;continue;}break;}cro2 = l.crosspoint(outer_line[f]);while(1) {if(l.linecrossseg(outer_line[t]) == 0) {t = (t + 1) % n;continue;}temp = l.crosspoint(outer_line[t]);if(sgn(P.distance(temp) - Q.distance(temp)) > 0) {t = (t + 1) % n;continue;}break;}cro1 = l.crosspoint(outer_line[t]);int near_i = t, near_i_1 = f;// cout << cro1.x << ' ' << cro1.y << '\n';
// cout << cro2.x << ' ' << cro2.y << '\n';double len = cro2.distance(outer[near_i_1]) + cro1.distance(outer[(near_i + 1) % n]);// cout << near_i << ' ' << near_i_1 << '\n';if(near_i_1 < near_i) {if(near_i_1 != 0) len += siz[(near_i_1 - 1 + n) % n];len += (siz[n - 1] - siz[near_i]);} else {len += siz[(near_i_1 - 1 + n) % n] - siz[near_i];}// cout << len << '\n';ans += l.length() * len;}printf("%.15f\n", ans / siz[n - 1]);
}signed main() {ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#ifdef ACM_LOCALfreopen("input", "r", stdin);freopen("output", "w", stdout);signed test_index_for_debug = 1;char acm_local_for_debug = 0;do {if (acm_local_for_debug == '$') exit(0);if (test_index_for_debug > 20)throw runtime_error("Check the stdin!!!");auto start_clock_for_debug = clock();solve();auto end_clock_for_debug = clock();cout << "Test " << test_index_for_debug << " successful!" << endl;cerr << "Test " << test_index_for_debug++ << " Run Time: "<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;cout << "--------------------------------------------------" << endl;} while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#elsesolve();
#endifreturn 0;
}
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