原题传送:http://poj.org/problem?id=3461 Oulipo Time Limit: 1000MS Memory Limit: 65536K Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from th

Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17795   Accepted: 7160 Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote

Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book: Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal,

首先先讲一下KMP算法作用: KMP就是来求在给出的一串字符(我们把它放在str字符数组里面)中求另外一个比str数组短的字符数组(我们叫它为ptr)在str中的出现位置或者是次数 这个出现的次数是可以重叠的 例如:在数组   ababa    中  aba的出现次数-----------它的答案是2------分别是从[0---2]  [2---4] 他们是可以重叠的,但是不可以完全重叠(废话T_T) KMP算法主要有两个模板 1.https://blog.csdn.net/starstar1

1.题目大意:单字符串匹配问题 2.分析:经典KMP问题 存个模板QAQ #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; char P[1000010]; char T[1000010]; int f[1000010]; inline void getfail(){ int m = strlen(T); f[0]

题意:给你两个字符串p和s,求出p在s中出现的次数. 显然,我们要先把模式串放到前面,之后主串放后面,中间隔开,这样就可以根据前缀数组的性质来求了. 我先想直接把p接到s前面,之后求Next数组对strlen(p)取余==0的就可以,之后WA.最后发现A AASSAAS的时候有bug,只有又想到在p和s中间加个不可能出现的字符'$'就可以了,戒指就A了. #include <bits/stdc++.h> using namespace std; const int INF=0x3f3f3f3f

这个算法去年的这个时候就已经听过了,看毛片算法哈哈..不过理解它确实花了我很久的时间..以致于我一直很排斥字符串的学习,因为总觉得太难了,但是有些硬骨头还是要啃的,这个寒假就啃啃字符串还有一些别的东西吧,KMP的学习我看了好多好多博客才有那么些头绪,复杂度的分析更是无从谈起,不过线性匹配这样的算法实在太流弊了.~题目是水题,但也算是我的第一道KMP吧.~ #include<iostream> #include<cstring> #include<string> #inc

题目大意 给定一个文本串和模式串,求模式串在文本串中出现的次数 题解 正宗KMP 代码: #include<iostream> #include<cstring> #include<string> #include<cstdio> using namespace std; #define MAXN 10005 ],P[MAXN]; int f[MAXN]; void getFail() { int m=strlen(P); f[]=f[]=; ; i<

# include <stdio.h> # include <algorithm> # include <string.h> using namespace std; char a1[1000010],a2[1000010]; int next[1000010]; int len1,len2,cot; void Getnext() { int i=0,j=-1; next[0]=-1; while(i<=len1) { if(j==-1||a1[i]==a1[j]

正解:kmp/哈希 解题报告: 传送门! 这题其实就kmp板子,,,用来复习下kmp的太久没打了QAQ 所以kmp做法就不港了放个代码就是了QAQ #include<algorithm> #include<iomanip> #include<cstring> #include<string> #include<cstdio> #include<cmath> using namespace std; #define il inline

题目链接:http://poj.org/problem?id=3461 代码如下: #include<cstdio>//poj 3461 kmp #include<cstring> using namespace std; char s1[1000005], s2[10005]; int len1, len2,next[10005]; void getnext() { //next数组其实是:当前字符匹配失败时,小字符串退回到合适的位置,然后继续匹配. int i = 0, j =

Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 36916   Accepted: 14904 Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quot

//poj3461 Oulipo //kmp模板 统计子串在母串中的位置 #include<iostream> #include<cstdio> #include<cstring> using namespace std; ]; ],s[]; int t,num,l1,l2; void get_next() { ; ; next[]=-; while(j<l2) { ||p[k]==p[j]) { k++; j++; next[j]=k; } else { k=n

先粘上我入门KMP时看的大佬的博客:orz orz 从头到尾彻底理解KMP 我觉得这篇已经讲的很详细了,希望大家能坚持看下去. 步骤 ①寻找前缀后缀最长公共元素长度对于P = p0 p1 ...pj-1 pj,寻找模式串P中长度最大且相等的前缀和后缀.如果存在p0 p1 ...pk-1 pk = pj- k pj-k+1...pj-1 pj,那么在包含pj的模式串中有最大长度为k+1的相同前缀后缀. 举个例子,如果给定的模式串为“abab”,那么它的各个子串的前缀后缀的公共元素的最大长度如下表格

一.需求:计算网页访问量前三名 import org.apache.spark.rdd.RDD import org.apache.spark.{SparkConf, SparkContext} /** * 需求:计算网页访问量前三名 * 用户:喜欢视频 直播 * 帮助企业做经营和决策 * * 看数据 */ object UrlCount { def main(args: Array[String]): Unit = { //1.加载数据 val conf:SparkConf = new Spa

Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 35694   Accepted: 14424 Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quot

http://poj.org/problem?id=3461 (题目链接) 题意 求一个字符串在另一个字符串中出现的次数. Solution KMP裸题,太久没写过了,都忘记怎么求next数组了..水一发→_→ 细节 next[0]=-1. 代码 // poj3461 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio>

Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book: Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal,

题面 The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book: Tout avait Pair normal, mais tout s'affirmait faux. Tout avait Fair normal, d'abord,

Oulipo Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 29759   Accepted: 11986 Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quot

http://poj.org/problem?id=3461 KMP板子,好久以前学过了,直接把板子粘上去即可. #include<cstdio> #include<cstring> using namespace std; ]; ]; ]={}; void getnext(int m){ ; ;i<=m;i++){ &&s2[j+]!=s2[i])j=nxt[j]; ]==s2[i])j++; nxt[i]=j; } return; } int ans; v