Trie模板.把所有单词都用字典树存起来,然后给每个节点赋权值表示前缀到该节点出现了几次.然后直接查询就可以了. #include <algorithm> #include <iostream> #include <sstream> #include <cstdlib> #include <climits> #include <cstring> #include <cstdio> #include <string&g…

题意:给n个字符串,给m个询问,每个询问给k个条形码.每个条形码由8个小码组成,每个小码有相应的宽度,已知一个条形码的宽度只有2种,宽的表示1,窄的表示0.并且宽的宽度是窄的宽度的2倍.由于扫描的时候有误差,每个小码的宽度为一个浮点型数据,保证每个数据的误差在5%内.所以一个条形码可以对应一个ASCC码,表示一个小写字母.k个条形码表示一个字符串s,每个询问表示给定的m个字符串中以s为前缀的字符串个数. 析:很容易看起来是Tire树, 不知道能不能暴过去,我觉得差不多可以,主要是在处理浮点数上,…

题意:给定n个字符串和m个经过处理得到的字符串,问对于m个字符串中的每个字符串,n个字符串中以该字符串为前缀的个数.分析:1.误差在[0.95x, 1.05x],因此求8个数的平均数,大于平均数为1,否则为0.2.字典树求前缀个数. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream&g…

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1780    Accepted Submission(s): 635 Problem Description After Gardon had got Angel's letter, he found it was encoded...Oh my god, why did she enco…

E - Encoded Barcodes Crawling in process...Crawling failedTime Limit:3000MS    Memory Limit:0KB    64bit IO Format:%lld & %llu SubmitStatusPracticeUVALive 5029 Description     All the big malls need a powerful system for the products retrieval. Now y…

B - Encoded Love-letter Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1591 Description After Gardon had got Angel's letter, he found it was encoded...Oh my god, why did she encode a love-lette…

F - City Game Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1505 Appoint description: Description Bob is a strategy game programming specialist. In his new city building game the gaming enviro…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1053 Entropy Description An entropy encoder is a data encoding method that achieves lossless data compression by encoding a message with “wasted” or “extra” information removed. In other words, entropy e…

Encoded Love-letter Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 35   Accepted Submission(s) : 14 Problem Description After Gardon had got Angel's letter, he found it was encoded...Oh my god,…

并不是很精简,随便改改A过了就没有再简化了. 1020. Problem Description Given a string containing only 'A' - 'Z', we could encode it using the following method: 1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only c…