题目传送门 /* 题意:从原点出发,四个方向,碰到一个点向右转,问多少次才能走出,若不能输出-1 模拟:碰到的点横坐标相等或纵坐标相等,然而要先满足碰到点最近, 当没有转向或走到之前走过的点结束循环.dir数组使得代码精简巧妙 对点离原点排序竟然submit failed,别人的代码有毒! */ #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include…

思路:离散化之后,直接模拟就行,标记vis开三维 代码: #include<iostream> #include<algorithm> #include<cstdio> #include<stdio.h> #include<string.h> #include<queue> #include<cmath> #include<map> #include<set> #include<vector&…

模拟. 每次找一下即将要遇到的那个点,这个数据范围可以暴力找,自己的写的时候二分了一下.如果步数大于$4*n$一定是$-1$. #include<bits/stdc++.h> using namespace std; const int INF = 0x7FFFFFFF; ; ; ; ; int T,n,m; struct P { int x,y; P(,) { x=X; y=Y; } }p[],q[]; bool cmp1(P a,P b) { if(a.x!=b.x) return a.x…

Right turn frog is trapped in a maze. The maze is infinitely large and divided into grids. It also consists of n obstacles, where the i-th obstacle lies in grid (xi,yi) . frog is initially in grid (0,0) , heading grid (1,0) . She moves according to T…

Text I was having dinner at a restaurant when Tony Steele came in. Tony worked in a layer's office years ago, but he is now working at a bank. He gets a good salary, but he always borrows money from his friends and never pays it back. Tony saw me and…

SCU 4437  Carries Time Limit:0MS     Memory Limit:0KB     64bit IO Format:%lld & %llu  Practice  Description Carries frog has nn integers a1,a2,…,ana1,a2,…,an, and she wants to add them pairwise. Unfortunately, frog is somehow afraid of carries (进位).…

 SCU 4438 Censor Time Limit:0MS     Memory Limit:0KB     64bit IO Format:%lld & %llu  Practice Description Censor frog is now a editor to censor so-called sensitive words (敏感词). She has a long text pp. Her job is relatively simple -- just to find the…

 SCU 4440 Rectangle Time Limit:0MS     Memory Limit:0KB     64bit IO Format:%lld & %llu  Practice Description Rectangle frog has a piece of paper divided into nn rows and mm columns. Today, she would like to draw a rectangle whose perimeter is not gr…

webRTC支持点对点通讯,但是webRTC仍然需要服务端:  . 协调通讯过程中客户端之间需要交换元数据,    如一个客户端找到另一个客户端以及通知另一个客户端开始通讯.  . 需要处理NAT(网络地址转换)或防火墙,这是公网上通讯首要处理的问题.    所以我们需要了解服务端相关的知识:信令.Stun.trun.ice. 一.什么是信令 信令就是协调通讯的过程,为了建立一个webRTC的通讯过程,客户端需要交换如下信息:  . 会话控制信息,用来开始和结束通话,即开始视频.结束视频这些操作…

原题:  Turn the pokers       思路:假设正面为0,反面为1.牌就像这样 000000....... .考虑到假如可以实现最终反面个数为m, 牌共n张, 则这n张排任取m个为反面其余都为正面的状况都能实现.于是转化为考虑最终可能出现1的个数的集合有哪些.       因为可能的个数集合是连续的(在最大最小值之内相差2的都可能), 所以每一次翻转之后的上下限都可以根据上一次所得的上下限推出.       最后算排列组合的适合需要用到组合数递推公式和费马小定理推论\( a^{p…