Description We all use cell phone today. And we must be familiar with the intelligent English input method on the cell phone. To be specific, the number buttons may correspond to some English letters respectively, as shown below: 2 : a, b, c    3 : d…

Keywords Search Description In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.       Wiskey also wants to bring this feature to his image retrieval system.       Every image have a long description, when users…

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4287 Intelligent IME Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2091    Accepted Submission(s): 1031…

Intelligent IME Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4776    Accepted Submission(s): 2227 Problem Description We all use cell phone today. And we must be familiar with the intelligen…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5536 题目大意是给了一个序列,求(si+sj)^sk的最大值. 首先n有1000,暴力理论上是不行的. 此外题目中说大数据只有10组,小数据最多n只有100.(那么c*n^2的复杂度应该差不多) 于是可以考虑枚举i和j,然后匹配k. 于是可以先把所有s[k]全部存进一个字典树, 然后枚举s[i]和s[j],由于i.j.k互不相等,于是先从字典树里面删掉s[i]和s[j],然后对s[i]+s[j]这个…

在我没用hash之前,一直TLE,字符串处理时间过长,用了hash之后一直CE,(请看下图)我自从经历我的字典树G++MLE,C++AC以后,一直天真的用C++,后来的CE就是因为这个,G++才支持这个hash... #include<cstdio>#include<iostream> #include<string.h> ]; struct TrieNode { int no; TrieNode *next[]; } node[]; TrieNode *root =…

Intelligent IME Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4287 Description We all use cell phone today. And we must be familiar with the intelligent English input method on the cell phone. To be specific,…

Intelligent IME Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1348    Accepted Submission(s): 685 Problem Description We all use cell phone today. And we must be familiar with the intelligent…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5512 学习菊苣的博客,只粘链接,不粘题目描述了. 题目大意就是给了初始的集合{a, b},然后取集合里的两个元素进行加或者减的操作,生成新的元素.问最后最多能生成多少个元素.问答案的奇偶性. 首先一开始有a, b.那么如果生成了b-a(b>a),自然原来的数同样可以由b-a, a生成(b != 2a). 于是如此反复下去,最后的数必然是可以由两个数p, 2p生成的. 于是所有的数肯定可以表示成xp+…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3915 题目大意是给了n个堆,然后去掉一些堆,使得先手变成必败局势. 首先这是个Nim博弈,必败局势是所有xor和为0. 那么自然变成了n个数里面取出一些数,使得xor和为0,求取法数. 首先由xor高斯消元得到一组向量基,但是这些向量基是无法表示0的. 所以要表示0,必须有若干0来表示,所以n-row就是消元结束后0的个数,那么2^(n-row)就是能组成0的种数. 对n==row特判一下. 代码:…