Oulipo poj-3461 题目大意:给你两个字符串s和p,问s中有多少个等于p的子串. 注释:$1\le strlen(p)\le 10^4\qquad1\le strlen(s)\le 10^6$ 想法:刚刚学习KMP,先来一道裸题.什么是KMP? KMP是一种字符串匹配算法.如果求出在一个母串中子串出现的位置,我们用一种办法就是枚举母串中任意一个位置pos[i]作为子串开始节点,然后向后匹配,如果匹配成功或者失配,我们都必须重新从pos[i]+1开始重新枚举.我们假设已经匹配到了p串的…

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>…

    KMP -- POJ3461解题报告 问题描述:给出字符串P和字符串T,问字符串P在字符串T中出现的次数 Sample Input 3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN Sample Output 1 3 0 简单KMP应用, 代码如下: //poj3461解题报告 #include<iostream> using namespace std; ], P[]; int len1, len2; ]; void getnext(char…

题目链接:https://vjudge.net/problem/POJ-3461 题意:给出主串和模式串,求出模式串在主串中出现的次数. 思路:kmp板子题. AC代码: #include<cstdio> #include<algorithm> #include<cstring> using namespace std; ; int T,next[maxn],len1,len2; char s1[maxn],s2[maxn]; //得到next数组 void get_n…

最近忙着考研复习,所以刷题少了.. 数据结构昨天重新学习了一下KMP算法,今天自己试着写了写,问题还不少,不过KMP算法总归是理解了,以前看v_JULY_v的博客,一头雾水,现在终于懂了他为什么要在算完next[]之后,所有值都减一了...... 为了顺便复习下C语言,就没有用C++写,而且写出来比较丑.. #include <string.h> #include <stdio.h> #include <stdlib.h> const int MAXSIZE = 100…

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…

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…

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]…

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,…

 题意:  计算一个字符串在另一个字符串中出现的次数. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; +; +; char t[maxl],p[maxw]; int T,ans,f[maxw]; void getfail(char *p) { int m=strlen(p); f[]=;f[]=; ;i<m;i++) { int j=f[i]; while(j…