Codeforce 633.C Spy Syndrome 2】的更多相关文章

C. Spy Syndrome 2 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such a…
题意:还是比较好理解 分析:把每个单词反转,建字典树,然后暴力匹配加密串 注:然后我就是特别不理解,上面那种能过,而且时间很短,但是我想反之亦然啊 我一开始写的是,把加密串进行反转,然后单词正着建字典树,然后就TLE了,反着写就能过 真是百思不得解,然后我猜测可能是单词数目比较少 #include <cstdio> #include <iostream> #include <cstring> #include <cstdlib> #include <v…
C. Spy Syndrome 2 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such a…
C. Spy Syndrome 2 题目连接: http://www.codeforces.com/contest/633/problem/C Description After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and a…
time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't u…
题目链接 C. Spy Syndrome 2 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy s…
Codeforces 633C Spy Syndrome 2 | Trie树裸题 一个由许多空格隔开的单词组成的字符串,进行了以下操作:把所有字符变成小写,把每个单词颠倒过来,然后去掉单词间的空格.已知操作后的字符串以及可能出现的所有单词,请输出原字符串(多组解只输出一个). 我犯的错误:数组的trie树--如果根节点是1--则其余节点要从2开始啊啊啊 #include <cstdio> #include <cstring> #include <algorithm> u…
Spy Syndrome 2 题意 现在对某个英文句子,进行加密: 把所有的字母变成小写字母 把所有的单词反过来 去掉单词之间的空格 比如:Kira is childish and he hates losing 加密为ariksihsidlihcdnaehsetahgnisol 现在给出加密后的句子,以及m个单词(每个单词可以重复使用),输出原来的句子. 思路 使用dp[i]表示第i个字母作为句子最后一个单词的开头所在的单词编号. dp值为0表示不能组成句子. 对于每个单词hash一下,按照每…
题目链接:http://codeforces.com/problemset/problem/633/C 大意就是给个字典和一个字符串,求一个用字典中的单词恰好构成字符串的匹配. 比赛的时候是用AC自动机写的,就是对于trie中每一个节点,判断是否为终结点,以及当前字符所在位置p减去trie中这个节点的深度也即某一单词的长度l,判断dp[p-l]是否可以被构成,可以的话直接break并且标记当前dp值. 赛后想了想,其实直接一个trie就行了,每个单词才1000的长度,又想起来我去年给人讲过类似的…
https://vjudge.net/problem/CodeForces-633C http://codeforces.com/problemset/problem/633/C 点击这里看巨佬题解 #include<iostream> #include<cstdio> #include<cstring> #include<vector> using namespace std; ; ,now,ll,pos[M]; char s[M],t[M]; vecto…