交互式程序,要用到一个函数fflush,它的作用是对标准输出流的清理,对stdout来说是及时地打印数据到屏幕上,一个事实:标准输出是以『行』为单位进行的,也即碰到\n才打印数据到屏幕.这就可能造成延时.在Windows平台上是看不出来的,它被改成及时生效了.而fflush对stdin的作用是清除冗余输入. #include<cstdio> ; //要AC ; bool vis[maxn][maxn]; ][maxlen] = {"NORTH","EAST&quo…

Problem B. Blind WalkTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86821#problem/B Description This is an interactive problem. Your task is to write a program that controls a robot which blindly wa…

Gym 101047E Escape from Ayutthaya Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u  Practice  Description standard input/output Ayutthaya was one of the first kingdoms in Thailand, spanning since its foundation in 1350 to…

Shopping 题目连接: http://codeforces.com/gym/100803/attachments Description Your friend will enjoy shopping. She will walk through a mall along a straight street, where N individual shops (numbered from 1 to N) are aligned at regular intervals. Each shop…

http://codeforces.com/gym/101147/problem/I I. On the way to the park time limit per test 5 seconds memory limit per test 64 megabytes input walk.in output standard output Engineers around the world share a lot of common characteristics. For example,…

os.walk()返回三个参数:os.walk(dirpath,dirnames,filenames) for dirpath,dirnames,filenames in os.walk(): 返回dirnames,filenames的类型为列表list[] 返回的dirpath为filenames的上级路径,如果要获得全路径和文件名,以便open的话,可以用在for filename in filenames的循环下用os.path.join(dirpath,filename)进行拼接…

 Gym 101047M Removing coins in Kem Kadrãn Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Practice Description standard input/output Andréh and his friend Andréas are board-game aficionados. They know many of their friend…

 Gym 101047K Training with Phuket's larvae Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Practice Description standard input/output Thai cuisine is known for combining seasonings so that every dish has flavors that are…

 Gym 101047B  Renzo and the palindromic decoration Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Practice Description standard input/output At the ruins of Wat Phra Si Sanphet (วดพระศรสรรเพชญ), one can find famous inscr…

/* 依旧考虑新增 2^20 个点. i 只需要向 i 去掉某一位的 1 的点连边. 这样一来图的边数就被压缩到了 20 · 2^20 + 2n + m,然后 BFS 求出 1 到每个点的最短路即可. 时间复杂度 O(20 · 2^20 + n + m) */ #include<cstdio> ,M=; int n,m,i,x,y,cnt,g0[N],g1[N],v[M],nxt[M],ed,h,t,q[N],d[N]; void add(int*g,int x,int y){v[++ed]=…