1071 Speech Patterns (25 分)   People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's id…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, which is useful when v…

一.技术总结 开始拿到这道题目时,思考的是我该如何区分它们每一个单词,不知道这里还是要学习得知在cctype头文件中有一个函数用于查看是否为0~9.a~z.A~Z,就是isalnum(),又因为题目中要求不区分大小写,有一个函数tolower(),toupper()要学会合理利用. 然后就是使用map技术了,默认初始化为0如果是string,int,可以直接mp[]++; 然后就是键值和值mp->first,mp->second; 二.参考代码 #include<iostream>…

Source: PAT A1071 Speech Patterns (25 分) Description: People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to na…

Java排序一,冒泡排序! 刚刚开始学习Java,但是比较有兴趣研究算法.最近看了一本算法笔记,刚开始只是打算随便看看,但是发现这本书非常不错,尤其是对排序算法,以及哈希函数的一些解释,让我非常的感兴趣,就记录一下自己的学习总结! 排序:将一些无序的元素按照某种规则排列的过程就叫"排序".在生活中,有时候可能是一些少量的数据 ,,,但是 ,也有可能是 一些的大数据 .排序是非常基础和重要的算法,有着广泛的理论基础和实践需求.(加粗部分摘自<算法笔记>原话!:-D) 一个排序…

1071. Speech Patterns (25) 时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 HOU, Qiming People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops&qu…

算法笔记 这个博客写的不错:http://blog.csdn.net/wust_zzwh/article/details/52100392 数位dp的精髓是不同情况下sta变量的设置. 模板: ]; ll dp[][state];//不同题目状态不同 ll dfs(int pos,/*state变量*/,bool lead/*前导零*/,bool limit/*数位上界变量*/)//不是每个题都要判断前导零 { //递归边界,既然是按位枚举,最低位是0,那么pos==-1说明这个数我枚举完了 )…

算法笔记 模板: vector<int>g[N]; vector<int>edge[N]; ][N]; int deep[N]; int h[N]; void dfs(int o,int u,int w) { ,h[u]=h[o]+w; ;j<g[u].size();j++) { if(g[u][j]!=o) { anc[][g[u][j]]=u; ;i<;i++)anc[i][g[u][j]]=anc[i-][anc[i-][g[u][j]]]; dfs(u,g[u]…

算法笔记 map: map<string,int> m; map<string,int>::iterator it;//auto it it = m.begin(); while(it != m.end()) { //it->first; //it->second; it ++; } set: multiset<int>ss; multiset<int>::iterator it=ss.begin();//auto it,multiset元素可以…