A - Inna and Choose Options Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Description There always is something to choose from! And now, instead of "Noughts and Crosses", Inna choose a very unusual upgrade of…

题意:给出n个点,分别属于k个集合,判断每个集合里面的点的距离都为0,为0的话输出yes,并输出任意两个集合之间的最短路 这道题目有两个地方不会处理, 先是n个点,分别属于k个集合,该怎么记录下来这里, 然后就是判断每个集合里面的点的距离是否为1,这里可以用并查集来做,如果在输入点的时候,距离为0,就将这两点合并 最后判断每个点,如果他们同属于一个集合,判断它俩的根是否一样就可以了 最后用floyd求最短路 #include<iostream> #include<cstdio> #…

题意:给出一个矩形的三种操作,顺时针旋转,逆时针旋转,对称,给出原始坐标,再给出操作数,问最后得到的坐标 画一下模拟一下操作就可以找到规律了 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #include<set> #inc…

题目链接:http://codeforces.com/contest/776/problem/D D. The Door Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Moriarty has trapped n people in n distinct rooms in a hotel. Some room…

A. A Serial Killer time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his power…

Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim. The killer starts with two potentia…

time limit per test4 seconds memory limit per test256 megabytes inputstandard input outputstandard output Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair li and ri and asks you to cou…

前四题比较水,E我看出是欧拉函数傻逼题,但我傻逼不会,百度了下开始学,最后在加时的时候A掉了 AC:ABCDE Rank:182 Rating:2193+34->2227 终于橙了,不知道能待几天 A.A Serial Killer 题目大意:一开始给你两个字符串,每次给你当前两个串中的一个和一个新的串,用新的串换掉旧的,每次输出当前的串.(次数<=1000) 思路:二逼题 #include<iostream> using namespace std; int main() { s…

最近好像总是有点不想打,专题也刷不动,还是坚持这做了一场,虽然打到一半就没打了...(反正通常都只能做出两题) 感觉自己切水题越来越熟练了,然而难题还是不会做.. A题,水,用vector存下来就行了 #include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #i…

再来回顾一下2-SAT,把每个点拆点为是和非两个点,如果a能一定推出非b,则a->非b,其他情况同理. 然后跑强连通分量分解,保证a和非a不在同一个分量里面. 这题由于你建完图发现都是双向边,所以用并查集亦可. #include<cstdio> #include<vector> #include<cstring> using namespace std; vector<int>G[200010],rG[200010],vs; bool used[200…