随笔分类 -  字符串

摘要:Family View Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 175 Accepted Submission(s): 20 Proble 阅读全文
posted @ 2016-09-17 21:01 Przz 阅读(813) 评论(2) 推荐(0)
摘要:Bazinga Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2287 Accepted Submission(s): 713 Problem 阅读全文
posted @ 2016-08-29 21:10 Przz 阅读(223) 评论(0) 推荐(0)
摘要:Boring String Problem Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1848 Accepted Submission(s) 阅读全文
posted @ 2016-08-21 09:59 Przz 阅读(3280) 评论(0) 推荐(0)
摘要:Boring counting Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2906 Accepted Submission(s): 1201 阅读全文
posted @ 2016-08-21 09:58 Przz 阅读(462) 评论(0) 推荐(0)
摘要:The Number of Palindromes Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2465 Accepted Submiss 阅读全文
posted @ 2016-08-21 09:55 Przz 阅读(314) 评论(0) 推荐(0)
摘要:Alice's Classified Message Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 312 Accepted Submis 阅读全文
posted @ 2016-04-06 22:33 Przz 阅读(232) 评论(0) 推荐(0)
摘要:后缀数组: sa[i]:表示排名第i个的首字母位置 Rank[i]:第i个数的排名 Height[i]:sa[i]和sa[i-1]的最长公共前缀 suffix(j) 和suffix(k) 的最长公共前缀为height[rank[j]+1], height[rank[j]+2], height[ran 阅读全文
posted @ 2016-04-06 22:25 Przz 阅读(1092) 评论(0) 推荐(0)
摘要:Power Strings Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 41110 Accepted: 17099 Description Given two strings a and b we define a*b to 阅读全文
posted @ 2016-04-06 22:18 Przz 阅读(185) 评论(0) 推荐(0)
摘要:Maximum repetition substring Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8669 Accepted: 2637 Description The repetition number of a str 阅读全文
posted @ 2016-04-06 22:16 Przz 阅读(229) 评论(0) 推荐(0)
摘要:Description You are the King of Byteland. Your agents have just intercepted a batch of encrypted enemy messages concerning the date of the planned att 阅读全文
posted @ 2016-04-06 22:13 Przz 阅读(276) 评论(0) 推荐(0)
摘要:Life Forms Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 12484 Accepted: 3502 Description You may have wondered why most extraterrestrial 阅读全文
posted @ 2016-04-06 22:10 Przz 阅读(291) 评论(0) 推荐(0)
摘要:Common Substrings Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 9248 Accepted: 3071 Description A substring of a string T is defined as: 阅读全文
posted @ 2016-04-06 21:53 Przz 阅读(175) 评论(0) 推荐(0)
摘要:Long Long Message Time Limit: 4000MS Memory Limit: 131072K Total Submissions: 25752 Accepted: 10483 Case Time Limit: 1000MS Description The little cat 阅读全文
posted @ 2016-04-06 21:48 Przz 阅读(180) 评论(0) 推荐(0)
摘要:1297. Palindrome Time limit: 1.0 secondMemory limit: 64 MB The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that t 阅读全文
posted @ 2016-04-06 21:46 Przz 阅读(354) 评论(0) 推荐(0)
摘要:694. Distinct Substrings Problem code: DISUBSTR Given a string, we need to find the total number of its distinct substrings. Input T- number of test c 阅读全文
posted @ 2016-04-06 21:45 Przz 阅读(183) 评论(0) 推荐(0)
摘要:Milk Patterns Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 13127 Accepted: 5842 Case Time Limit: 2000MS Description Farmer John has noti 阅读全文
posted @ 2016-04-06 21:42 Przz 阅读(279) 评论(0) 推荐(0)
摘要:Musical Theme Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 24507 Accepted: 8259 Description A musical melody is represented as a sequenc 阅读全文
posted @ 2016-04-06 21:40 Przz 阅读(153) 评论(0) 推荐(0)
摘要:稍微学习了下第一次用后缀数组- - , 强行凑出答案 , 感觉现在最大的问题是很多算法都不知道 ,导致有的题一点头绪都没有(就像本题)。 /*推荐 《后缀数组——处理字符串的有力工具》——罗穗骞 后缀数组sa, sa[ i ] = a表示字符串从第a个开始到结尾的字典序排序为i 本题是绕成了一个环, 阅读全文
posted @ 2015-09-20 23:25 Przz 阅读(213) 评论(0) 推荐(0)
摘要:Problem Description CRB has two strings s and t. In each step, CRB can select arbitrary character c of s and insert any character d (d ≠ c) just after 阅读全文
posted @ 2015-08-31 03:15 Przz 阅读(208) 评论(0) 推荐(0)
摘要:Sample Input 2 abc abaadada 2 abc abaadada Sample Output Yes No Yes No 判断是否能成为3个非空回文子串 manacher算法求出个点回文长度,在找出第一个和最后一个保存下来,再判断中间的 阅读全文
posted @ 2015-08-17 15:34 Przz 阅读(253) 评论(0) 推荐(0)