Compound Words You are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is theconcatenation of exactly two other words in the dictionary. Input Standard input consists of a number of l…

Problem E: Compound Words You are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary. Input Standard input consists of a…

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1332 题目大意: 给定一个词典(已经按照字典序排好),要求找出其中所有的复合词,即恰好由两个单词连接而成的单词.(按字典序输出) 思路: 对于每个单词,存入Hash表,然后对每个单词拆分. Hash函数的选取可以看:https://www.byvoid.com/blog/string-has…

You are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.InputStandard input consists of a number of lowercase words, o…

  You are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary. Input   Standard input consists of a number of lowercase w…

今天下午略感无聊啊,切点水题打发打发时间,=_=|| 把所有字符串插入到一个set中去,然后对于每个字符串S,枚举所有可能的拆分组合S = A + B,看看A和B是否都在set中,是的话说明S就是一个复合词. #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <string> #include <set> #in…

题目 题目     分析 自认已经很简洁了,虽说牺牲了一些效率     代码 #include <bits/stdc++.h> using namespace std; set <string> m; string s[120003]; int main() { int n; while(cin>>s[n]) m.insert(s[n++]); for(int i=0;i<n;i++) { int l=s[i].length(); for(int j=1;j<…

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1332 #include<iostream> #include<stdio.h> #include<string.h> #include<string> #include<map> using namespace std; stri…

这个题,单纯做出来有很多种方法,但是时间限制3000ms,因此被TL了不知道多少次,关键还是找对最优解决方法,代码附上: #include<bits/stdc++.h> using namespace std; map<string,int>MAP; ][]; int main(){ MAP.clear(); ; while(cin>>s[n]){ MAP[s[n]]=; n++; } ;i<n;i++){ int l=strlen(s[i]); ;j<l;…

题目描述: 题目思路: 用map保存所有单词赋键值1,拆分单词,用map检查是否都为1,即为复合词 #include <iostream> #include <string> #include <map> using namespace std; map<string,int> dict ; ] ; int main(int argc, char *argv[]) { ; while(cin >> str[count]){ dict[str[co…