Seek the Name, Seek the Fame(Kmp)】的更多相关文章

Seek the Name, Seek the Fame Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 12   Accepted Submission(s) : 7 Problem Description The little cat is so famous, that many couples tramp over hill an…
Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11831   Accepted: 5796 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names t…
题意:给一个字符串s,问s的某个前缀与后缀相同的情况时,长度是多少. 此题使用KMP的next数组解决. next数组中,j=next[i],next[i]表示S[0...i-1]的某个后缀(字符串S[i-j,i-1])与字符串S[0...j-1]完全相同,此时的j即该段字符串的长度. 字符串s本身是最长的串,next[L]对应的S[0...next[L]]是次长的符合条件的串,其他更短的串怎么求呢. 由于字符串S[L-j...L-1]等于S[0...j-1],这样在S[0...j-1]时继续利…
poj2406 Power Strings(kmp) 给出一个字符串,问这个字符串是一个字符串重复几次.要求最大化重复次数. 若当前字符串为S,用kmp匹配'\0'+S和S即可. #include <cstdio> #include <cstring> using namespace std; const int maxn=2e6+5; char s1[maxn], s2[maxn]; int n1, n2, nxt[maxn], ans; int main(){ while (~…
http://poj.org/problem?id=2752 Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14611   Accepted: 7320 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked…
Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13840   Accepted: 6887 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names t…
Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24077   Accepted: 12558 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names…
Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative inte…
题意 给定一个字符串 \(S\) ,一次操作可以在这个字符串的右边增加任意一个字符.求操作之后的最短字符串,满足操作结束后的字符串是回文. \(1 \leq |S| \leq 10^6\) 思路 \(\text{KMP}\) 的 \(fail\) 数组是整个算法最重要的东西,能拓展出很多东西. 对于一个模式串(pattern)\(P\) ,\(fail\) 数组为 \(f\) ,那么 \(f[i]\) 就是如果匹配到 \(i\) 这个位置失配,下次去尝试的位置,也就说明了从字符串头到 \(f[i…
题目大意:S(n,k)用k(2-16)进制表示1-n的数字所组成的字符串,例如S(16,16)=123456789ABCDEF10: 解题思路: n最大50000,k最大100000,以为暴力会超时.但确实可以暴力(wtf),直接将给的十进制数表示为K进制数.再暴力进行字符串比较即可(string的find函数??)或者使用KMP算法: 不确定A不AC的代码(其实是AC的): #include<bits/stdc++.h> using namespace std; ]; void getfil…