Acesrc and String Theory Problem Description Acesrc is a famous string theorist at Nanjing University second to none. He insists that we should always say an important thing k times. He also believes that every string that can be obtained by concaten…

1.修改启动参数:打开[SQL Server 配置管理器(SQL Server Configuration Manager)]→右键[SQL Server(MSSQLSERVER)]属性→高级(Advanced)选项→双击[启动参数(Start Parameters)]→在文本末尾新增 ;-m →确定(OK) 2.重启SQL Server服务,如果服务不能启动,则更改登录身份为Local System 3.打开[SQL Server Management Studio]使用[Windows 身份…

557. Reverse Words in a String III[easy] Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. Example 1: Input: "Let's take LeetCode contest" Output:…

大意: 求重复$k$次的子串个数 枚举重复长度$i$, 把整个串分为$n/i$块, 如果每块可以$O(1)$计算, 那么最终复杂度就为$O(nlogn)$ 有个结论是: 以$j$开头的子串重复次数最大为$1+\lfloor\frac{lcp(j,j+i)}{i}\rfloor$ 先特判掉$k=1$的情况, 然后枚举每个块开头的位置, 计算出$lcp$的值$p$, 由于$k>1$, 合法位置的$lcp$值至少要跨越一个块, 可以得到 若$p\ge ki-1$, 那么这个块内所有点都合法 若$k(i…

String Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1512    Accepted Submission(s): 668 Problem Description Give you a string with length N, you can generate N strings by left shifts…

一.int转string 1.c++11标准增加了全局函数std::to_string: string to_string (int val); string to_string (long val); string to_string (long long val); string to_string (unsigned val); string to_string (unsigned long val); string to_string (unsigned long long val);…

Implement a method to perform basic string compression using the counts of repeated characters. For example, the string aabcccccaaa would become a2b1c5a3. If the "compressed" string would not become smaller than the original string, your method…

String类 1.String对象的初始化 由于String对象特别常用,所以在对String对象进行初始化时,Java提供了一种简化的特殊语法,格式如下: String s = “abc”; s = “Java语言”; 其实按照面向对象的标准语法,其格式应该为: String s = new String(“abc”); s = new String(“Java语言”); 只是按照面向对象的标准语法,在内存使用上存在比较大的浪费.例如String s = new String(“abc”);…

先求SA,然后按字典序从小到大枚举子串,每到一个后缀从长到短枚举子串(跳过长为he[i]的和前一段重复的子串),然后维护一个点p,保证i~p之间最小的he>=当前枚举长度,p是单调向右移的 然后把每个后缀的结果倒着输出即可 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=3005; int n,m,wa[N],wb[N],wv[N],wsu…

把两个串接到一起求一个SA,然后找最大的sa[i]和sa[i-1]不是一个串的he[i] #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=500005; int n,m,wa[N],wb[N],wv[N],wsu[N],sa[N],rk[N],he[N],bl[N],ans; char s[N],t[N]; bool cmp(int r[],in…