Word Amalgamation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9794   Accepted: 4701 Description In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but…

传送门:ZOJ1181  思路:自身排序来判断两个字符串拥有相同的字符.   #include<cstdio> #include<cstdlib> #include<iostream> #include<cstring> #include<string> #include<cmath> #include<memory.h> #include<algorithm> using namespace std; str…

Word Amalgamation Time Limit: 1 Sec  Memory Limit: 64 MB Submit: 373  Solved: 247 Description In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but in order to find th…

Word Amalgamation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2586    Accepted Submission(s): 1246 Problem Description In millions of newspapers across the United States there is a word game…

 Word Amalgamation  In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but in order to find the letters that appear in the answer it is necessary to unscramble four wor…

版权声明:本文为博主原创文章.未经博主同意不得转载. vasttian https://blog.csdn.net/u012860063/article/details/35338617 转载请注明出处:http://blog.csdn.net/u012860063天资 题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=1113 postid=17794&messageid=1&deep=0" rel="nofollow&q…

##统计word中的各个字符的出现的次数,并统计出所有前十名的字符使用次数 # -*- coding:utf-8 -*- word='''awfesdafhjkcasadckjsdackjsadvcnksausafdsch fsadfdsaasdfsdacsafsdaas csaasfdasdfsda sfadfsdafsadfjtyurjryjghnkuitki''' list1 = list(word) #将字符串转化为列表 while '\n' in list1: #去掉列表中的'\n'字…

有时,我们会碰到对字符串的排序,若采用一些经典的排序算法,则时间复杂度一般为O(n*lgn),但若采用Trie树,则时间复杂度仅为O(n). Trie树又名字典树,从字面意思即可理解,这种树的结构像英文字典一样,相邻的单词一般前缀相同,之所以时间复杂度低,是因为其采用了以空间换取时间的策略. 下图为一个针对字符串排序的Trie树(我们假设在这里字符串都是小写字母),每个结点有26个分支,每个分支代表一个字母,结点存放的是从root节点到达此结点的路经上的字符组成的字符串. 将每个字符串插入到tr…

Word Amalgamation 点我挑战题目 点我一起学习STL-MAP 题意分析 给出字典.之后给出一系列======乱序======单词,要求你查字典,如过这个乱序单词对用有多个有序单词可以输出,那么按照字典序将其输出. 若没有对应单词,输出NOT A VALID WORD. 可见这是一组组对应关系,可以用map来实现.map字典中first保存原本的单词(因为first按字典序),second保存原本单词排序后的单词.每次读入一个乱序单词后,sort遍历map并查找和second匹配的…

http://acm.hdu.edu.cn/showproblem.php?pid=1113 给定一个字典,然后每次输入一个字符串问字典中是否有单词与给定的字符串的所有字母一样(顺序可以打乱),按字典序输出字典中的原字符串. 我开始是直接用了 sort, 用一个结构体记录了所有字符串,和相应下标,输出的时候在用了冒泡排序. #include <iostream> #include <cstdio> #include <cmath> #include <vector…