以贝茜为参照物,则贝茜固定于原点,每个杀手是一个圆心在某条射线上的圆. 解出每个杀手可以射杀贝茜的时间区间,然后扫描线即可,时间复杂度$O(n\log n)$. #include<cstdio> #include<algorithm> #include<cmath> using namespace std; typedef long long ll; int n,r,bx,by,bvx,bvy,x,y,vx,vy,i,m,t,ans;double L,R,inf=1e9…
传送门 线段树区间修改傻题 #include <cstdio> #include <cstring> #include <iostream> #define N 50001 #define root 1, 1, n #define ls now << 1, l, mid #define rs now << 1 | 1, mid + 1, r int n, m, q, res; int ans[16][N << 2], sum[16][…
Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…
网络时延的构成 Network delay including four parts: Processing delay - time routers take to process the packet header. Queuing delay - time the packet spends in routing queues Transmission delay - time it takes to push the packet's bits onto the link. Propag…