HDU 1627 Krypton Factor】的更多相关文章

http://acm.hdu.edu.cn/showproblem.php?pid=1627 给定 n 和 L 找出第n个范围在0-L之内的字符串,字符串要求没有相邻的子串是相同的. 按照格式输出. 这个题的关键在于判断字符串是否是符合要求的字符串. 枚举字符串所有子串可能的长度,然后判断即可. #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include…
回溯法:避免无用判断,强化回溯代码的实现过程 题目的大意就是以字典序为准,排列字符串,但要保证一个字符串中不包含相邻的重复子串. Problem Description For example, the sequence ABACBCBAD is easy, since it contains an adjoining repetition of the subsequence CB. Other examples of easy sequences are: BB ABCDACABCAB AB…
Krypton Factor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 181    Accepted Submission(s): 60 Problem Description You have been employed by the organisers of a Super Krypton Factor Contest in…
UVA.129 Krypton Factor (搜索+暴力) 题意分析 搜索的策略是:优先找长串,若长串不合法,则回溯,继续找到合法串,直到找到所求合法串的编号,输出即可. 注意的地方就是合法串的判断,根据后缀的规则来判断,枚举后缀长度[1,len/2],后缀中是否有重复子串,若是的话表明不是合法串. 还有一个注意的地方,每次递归调用时,序号就要+1,无论是回溯回来的递归,还是深度搜索的递归,因为没找到一组可行解,编号就要加一.原因是此题不是用递归深度来判断输出的. 代码总览 #include…
/*UVa129 - Krypton Factor --回溯问题.看例子可知道确定该字符串是按照从左到右依次考虑每个位置,当前位置填不上所有的字符时,需要回溯. -- */ #define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<fstream> #include<time.h> #include<vector> using namespace std; const int maxn = 10…
 Krypton Factor Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 392    Accepted Submission(s): 174 Problem Description You have been employed by the organisers of a Super Krypton Factor Conte…
The Factor  Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=628&pid=1001 Description 有一个数列,FancyCoder沉迷于研究这个数列的乘积相关问题,但是它们的乘积往往非常大.幸运的是,FancyCoder只需要找到这个巨大乘积的最小的满足如下规则的因子:这个因子包含大于两个因子(包括…
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5428 The Factor  Accepts: 101  Submissions: 811  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 有一个数列,FancyCoder沉迷于研究这个数列的乘积相关问题,但是它们的乘积往往非常大.幸运的是,FancyCoder只需要找到…
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Problem Description There is a sequence of n positive integers. Fancycoder is addicted to learn their product, but this product may be extremely huge! However, it is lucky that FancyCoder only needs to find out one factor of this huge product: the sm…