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Java大数模板 二、Java之输入输出处理由于ACM竞赛题目的输入数据和输出数据一般有多组（不定），并且格式多种多样，所以，如何处理题目的输入输出是对大家的一项最基本的要求。这也是困扰初学者的一大问题。 ...

J .Distance to work 题意 懒得看了 题解 计算几何。 扒一扒dls的板子 typedef const inline inline struct db x, y; P() {} P(db _x, db _y) : x(_x), y(_y) {} P operator P operator P operator P operator bool int if return } bool return } db dot db det db distTo db alpha void void db abs db abs2 P rot90 P unit int P rot }; struct P ps[2 L() {} L(P p1,P p2) { ps[0 P& operator P dir bool L push const P delta = (ps[1 return } }; #define #define bool db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2); return } P isLL db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2); return } P isLL bool if return } bool return crossOp(p1,p2,q1) * crossOp(p1,p2,q2) 0 } return } vector int sort(ps.begin(), ps.end()); vector for while for while qs.resize(k - 1 return } vector //caution: need to unique the Ps first int sort(ps.begin(), ps.end()); vector for while for while qs.resize(k - 1 return } db convexDiameter int int int db ret = ps[i].distTo(ps[j]); do if (++j)%=n; else (++i)%=n; ret = max(ret,ps[i].distTo(ps[j])); }while return } vector vector int rep(i,0 P p1 = ps[i], p2 = ps[(i+1 int if if } return } //min_dist db min_dist if db ret = 1e100 rep(i,l,r) rep(j,l,i) ret = min(ret,ps[i].distTo(ps[j])); return } int db ret = min(min_dist(ps,l,m),min_dist(ps,m,r)); vector sort(qs.begin(), qs.end(),[](P a,P b) -> bool rep(i,1 ret = min(ret,qs[i].distTo(qs[j])); return } int db d = o1.distTo(o2); if if if if return } vector db x = (p1-o).dot(p2-p1), y = (p2-p1).abs2(), d = x * x - y * ((p1-o).abs2() - r*r); if d = max(d,0.0 return } vector db d = o1.distTo(o2); if if d = min(d, r1 + r2); db y = (r1 * r1 + d * d - r2 * r2) / (2 P dr = (o2 - o1).unit(); P q1 = o1 + dr * y, q2 = dr.rot90() * x; return } vector db x = (p - o).abs2(), d = x - r * r; if P q1 = o + (p - o) * (r * r / x); P q2 = (p - o).rot90() * (r * sqrt return } vector vector if P dr = (o2 - o1).unit().rot90() * r1; ret.pb(L(o1 + dr, o2 + dr)), ret.pb(L(o1 - dr, o2 - dr)); } else P p = (o2 * r1 - o1 * r2) / (r1 - r2); vector rep(i,0 } return } vector vector P p = (o1 * r2 + o2 * r1) / (r1 + r2); vector rep(i,0 return } db areaCT vector if bool if if sign((p1-is[0 return else } if if return } bool bool bool if return } else return } } bool if return } else return } } bool return } vector sort(l.begin(), l.end()); deque for if while while q.push_back(l[i]); } while while vector for return } P inCenter double return } P circumCenter P bb = b - a, cc = c - a; double return } P othroCenter P ba = b - a, ca = c - a, bc = b - c; double A = ca.x * ba.y - ba.x * ca.y, x0 = (Y + ca.x * ba.y * b.x - ba.x * ca.y * c.x) / A, y0 = -ba.x * (x0 - c.x) / ba.y + ca.y; return } AC代码（别人家的模板。） #include using #define const const const const // Compares a double to zero int if if return else return } // square of a double inline return } /* 15 * Point 16 * Point() - Empty constructor 17 * Point(double _x,double _y) - constructor 18 * input() - double input 19 * output() - %.2f output 20 * operator == - compares x and y 21 * operator 作者 CheaSim 发布于 2018-09-05 更新于 2018-09-05 许可协议 #[计算几何](/tags/%E8%AE%A1%E7%AE%97%E5%87%A0%E4%BD%95/)

回文树学习笔记 神奇的数据结构 Palindromic Tree 回文树 功能简介 求一个串$S$中$[0,i]$中本质不同回文串的个数 求串$S$中每一个本质不同回文串出现的次数 求指定下标$i$结尾的回文串的个数 变量简介 $len[i]$表示编号为$i$节点所表示的回文串的$len$。 ...