Find the Clones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 6365   Accepted: 2375 Description Doubleville, a small town in Texas, was attacked by the aliens. They have abducted some of the residents and taken them to the a spaceship…

Apache Tomcat 9 is not available from the standard RHEL distributions, so this article provides information about the manual installation and basic configuration of Apache Tomcat 9 on RHEL and its clones from tarballs. The tarball approach to install…

题目:Problem D. Clones and TreasuresInput file: standard inputOutput file: standard outputTime limit: 1 secondMemory limit: 256 mebibytesThe magical treasury consists of n sequential rooms. Due to construction of treasury its impossible togo from room…

Find the Clones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 8490   Accepted: 3210 Description Doubleville, a small town in Texas, was attacked by the aliens. They have abducted some of the residents and taken them to the a spaceship…

Abstract 大数据的克隆检测和搜索算法已经作为嵌入在应用中的一部分. 本文推出一个代码检测基准.包含一些已知的真假克隆代码.其中包括600万条真克隆(包含type-1,type-2,type-3,type-4). S1   Introduction IJaDataset 2.0包含25,000个系统,365MLOC(Million Lines of Code).通过人工标记找出实现相似功能的代码,10个功能,包含6百万真克隆对,26万假克隆对.其他的基准都不存在对语义克隆的检测. S2 b…

链接:http://poj.org/problem?id=2945 Description Doubleville, a small town in Texas, was attacked by the aliens. They have abducted some of the residents and taken them to the a spaceship orbiting around earth. After some (quite unpleasant) human experi…

https://vjudge.net/problem/POJ-2945 题意: 给出n个长度相同的DNA序列,如果一个DNA序列出现过两次,那么就有说明它被复制了一次.问被复制0次,1次,2次--n-1次的DNA序列分别有多少个. 思路: 可以利用字典树的方法做,用map目测会超时.因为一个字符串是它自己本身的假前缀,所以把每一个字符串的前缀数量-1,就是它被复制的数目.这题学到了一个技巧,统计的时候,已经统计过的要跳开,一开始用map判断超时了,之后想到用sort对char类型的字符串进行排序…

http://poj.org/problem?id=2945 还是trie树……对于结束标记累加并且开个数组记录一下即可. #include<cstdio> #include<cstring> using namespace std; struct node{ int ed; ]; void clear(){ memset(son,,sizeof(son)); ed=; } }tree[]; inline int turn(char ch){ ; ; ; ; } ][]; ]; i…

给你一行房间,有的是隐身药水,有的是守卫,有的是金币. 你可以任选起点,向右走,每经过一个药水或金币就拿走,每经过一个守卫必须消耗1个药水,问你最多得几个金币. 药水看成左括号,守卫看成右括号, 就从每个位置贪心地向右走,如果在 r 遇到不匹配,则把下一次的左端点置成r+1,接着走. O(n)即可. 因为如果把左端点放在上次的l和r之间,要么会发生不匹配,要么答案无法比上次走的更优. 队友代码: #include <iostream> #include <cstdio> #incl…

建一颗$trie$树(当然你哈希也资瓷),边插边更新,看看搜到最底时有多少个字符串,然后更新. #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<cctype> #include<cstdlib> #include<vector> #include<queue…