Crazy Search

题目: 链接:点击打开链接 题意: 输入n和nc,以及字符串s,输出长度为n的不同字串的个数. 算法: 思路: 用hash判重(hash值......),看了大牛的代码,对hash还是不甚理解.... 代码: #include<iostream> #include<cstring> #include<cstdio> #include<cstdio> using namespace std; #define MAXN 16000010 const int MA

Description Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a given text. Such number could be the number of different substrings of a given size that exist in t

Crazy Search Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 27336   Accepted: 7641 Description Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a gi

Crazy Search Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 26713 Accepted: 7449 Description Many people like to solve hard puzzles some of which may lead them to madness. One such puzzle could be finding a hidden prime number in a given

题目大意: 分析长度为n的子串有多少种. 思路分析: 对于没出现的字符,将其分配一个数字. 然后将子串看做一个nc进制的数. 然后hash判断. #include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <cstring> #include <string> using namespace std; bool vis[26666

题目意思: 给出一个字符串和字串的长度,求出该字符串的所有给定长度的字串的个数(不相同). 题目分析: 此题为简单的字符串哈hash map问题,可以直接调用STL里的map类.map<string,int> snum; AC代码: #include<iostream> #include<string> #include<map> using namespace std; int main() { int t,n,nc; cin>>t; whil

Problem地址:http://acm.hdu.edu.cn/showproblem.php?pid=1381 此题可以用哈希数组求解,至于什么是哈希,可以先看以下这篇推荐文章,写得挺不错的. 推荐:http://www.cnblogs.com/yangecnu/p/Introduce-Hashtable.html#undefined 首先是求key值,我采取了,求余法. key = value mod p 接着是确定value与p的值. p的值根据可能的组合数进行确定. 而value的值,为

Poj 1200 题意:给你一个n和m以及一个有m个不同字母组成的字符串,问有多少个长度为n的不同字符子串: 题解:以m为进制进行Hash.虽然是看了解题报告才会的但必须要理解并且学会运用:https://www.cnblogs.com/gj-Acit/archive/2013/05/15/3080734.html. #include<cstring>//别用set,set的添加是红黑树实现时间复杂度是O(logN),在这里会超时 #include<cstdio> #define

前置芝士 :string 的 基本用法 string s = "hello world" ; string tmp(s,0,5) ; cout << tmp << endl ; 上面这一段代码 可以复制粘贴 然后改变 数字.(试试效果 \(string\ tmp(s,i,n);\) 赋初值 也就是说从 $s[i]?- To - ?s[i+n-1] $ 所以既然是这样 应该就有一种比较优秀的做法 \(O(S.length())\) 可海星. \[STL大法好啊\]