题库链接http://acm.hdu.edu.cn/showproblem.php?pid=3336 这道题是KMP的next数组的一个简单使用,首先要理解next数组的现实意义:next[i]表示模式串的前i个字符所组成的字符串的最长前缀后缀匹配长度,就比如对于字符串"abcdabe",它的next[3]=0,因为前三个字符构成的字符子串"abc"前后缀最长匹配长度为0,而next[6]=2,因为对于子串"abcdab",它的前缀后缀最大可以匹…

题目链接:https://vjudge.net/problem/HDU-3336 Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11760    Accepted Submission(s): 5479 Problem Description It is well known that AekdyCoi…

Count the string Problem Description It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:s: "abab"The prefi…

D - Count the string Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can wr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3336 很容易想到用kmp 这里是next数组的应用 定义dp[i]表示以s[i]结尾的前缀的总数 那么dp[i]=dp[next[i]]+1; 代码: #include<stdio.h> #include<string.h> ; ; int dp[MAXN]; char str[MAXN]; int next[MAXN]; void getNext(char *p) { int j,k…

欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - HDU3336 题意概括 给T组数据,每组数据给一个长度为n的字符串s.求字符串每个前缀出现的次数和,结果mod 10007. 题解 首先闭着眼睛KMP跑一跑. 然后我们来dp. dp[i]表示以第i位结尾的前缀个数. 那么,根据Next的含义,不难写出dp[i]=dp[Next[i]]+1的转移方程式. 然后就OK了. 代码 #include <cstring> #include <algorit…

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:s: "abab"The prefixes are: "a", "ab"…

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:  s: "abab"  The prefixes are: "a", "ab&…

问题描述众所周知,aekdycoin擅长字符串问题和数论问题.当给定一个字符串s时,我们可以写下该字符串的所有非空前缀.例如:S:“ABAB”前缀是:“A”.“AB”.“ABA”.“ABAB”对于每个前缀,我们可以计算它在s中匹配的次数,因此我们可以看到前缀“a”匹配两次,“ab”也匹配两次,“ab a”匹配一次,“ab ab”匹配一次.现在,您需要计算所有前缀的匹配时间之和.对于“abab”,它是2+2+1+1=6.答案可能非常大,因此输出答案mod 10007. 输入第一行是一个整数t,表示…

这道题本来想对了,可是因为hdu对pascal语言的限制是我认为自己想错了,结果一看题解发现自己对了…… 题意:给以字符串 计算出以前i个字符为前缀的字符中 在主串中出现的次数和 如: num(abab)=num(a)+num(ab)+num(aba)+num(abab)=2+2+1+1=6; 题解:next[i]记录的是 长度为i 不为自身的最大首尾重复子串长度  num[i]记录长度为next[i]的前缀所重复出现的次数 推介一篇博文,非常不错,和本代码解法不一样,但实质上是一样的. 附上代…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=3336 如果你是ACMer,那么请点击看下 题意:求每一个的前缀在母串中出现次数的总和. AC代码: #include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include…

题意: 求一个字符串的所有前缀串的匹配次数之和. 思路: 首先仔细思考: 前缀串匹配. n个位置, 以每一个位置为结尾, 就可以得到对应的一个前缀串. 对于一个前缀串, 我们需要计算它的匹配次数. k = next [ j ] 表示前缀串 Sj 的范围内(可以视为较小规模的子问题), 前缀串 Sk 是最长的&能够匹配两次的前缀串. 这和我们需要的答案有什么关系呢? 题目是求所有前缀串的匹配次数之和, 那么可以先求前缀串 Si 在整个串中的匹配次数, 再加和. 到此, 用到了两个"分治&q…

dp[i]代表前i个字符组成的串中所有前缀出现的次数. dp[i] = dp[next[i]] + 1; 因为next函数的含义是str[1]~str[ next[i] ]等于str[ len-next[i]+1 ]~str[len],即串的前缀后缀中最长的公共长度. 对于串ababa,所有前缀为:a, ab,aba,abab, ababa, dp[3] = 3; 到达dp[5]的时候,next = 3, 它与前面的最长公共前缀为aba,因此dp[5]的凑法应该加上dp[3],再+1是加上aba…

题目链接 #include <bits/stdc++.h> using namespace std; typedef long long ll; inline int read() { ,f=;char ch=getchar(); ;ch=getchar();} +ch-';ch=getchar();} return x*f; } /********************************************************************/ ; char s[ma…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3797    Accepted Submission(s): 1776 Problem Description It is well known that AekdyCoin is good at string problems as well as nu…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3351    Accepted Submission(s): 1564 Problem Description It is well known that AekdyCoin is good at string problems as well as n…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8845    Accepted Submission(s): 4104 Problem Description It is well known that AekdyCoin is good at string problems as well as n…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6062    Accepted Submission(s): 2810 Problem Description It is well known that AekdyCoin is good at string problems as well as nu…

http://acm.hdu.edu.cn/showproblem.php?pid=3336 Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6672    Accepted Submission(s): 3089 Problem Description It is well known that Aek…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3637    Accepted Submission(s): 1689 Problem Description It is well known that AekdyCoin is good at string problems as well as nu…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4105    Accepted Submission(s): 1904 Problem Description It is well known that AekdyCoin is good at string problems as well as nu…

Count the string Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14096    Accepted Submission(s): 6462 Problem Description It is well known that AekdyCoin is good at string problems as well as…

Count New String 题意: 定义字符串函数 \(f(S,x,y)(1\le x\le y\le n)\),返回一个长度为y-x+1的字符串,第 i 位是 \(max_{i=x...x+k-1}S_i\) 设集合\(A = {f(f(S, x_1,y_1),x_2-x_1+1,y_2-x_1+1)|1\le x_1 \le x_2 \le y_2 \le y_2 \le n}\) 求集合A 的大小 \(N\le 1e5\) 字符集大小 <=10 分析: 先放出官方题解 方法一 核心点…

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:s: "abab"The prefixes are: "a", "ab"…

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example: s: "abab" The prefixes are: "a", "ab&qu…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3336 题目大意:找出字符串s中和s的前缀相同的所有子串的个数. 题目分析:KMP模板题.这道题考虑 nxt[] 数组的应用.以 s[i] 结尾的子串中一共有多少个子串可以作为s的前缀呢?我们只要令 t = nxt[i],cnt=0 每当 t!=-1,cnt++, t=nxt[t] 就可以了. 当然,我们可以用dp优化,dp[i] = dp[nxt[i]]+1 ,当然,如果 nxt[i]==-1 ,那…

题意 计算所有S的前缀在S中出现了几次 思路 跟前缀有关的题目可以多多考虑KMP的NEXT数组 #include <cstdio> #include <cstring> #include <iostream> #include <cstdlib> using namespace std; char S[2000000]; int NEXT[2000000]; int dp[2000000];//dp[i] 表示 以i结尾的子串与前缀相等的个数 d[i]=d[…

KMP算法的综合练习 DP很久没写搞了半天才明白.本题结合Next[]的意义以及动态规划考察对KMP算法的掌握. Problem Description It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this str…

居然一A了,说明对朴素的KMP还是有一定理解. 主要就是要知道next数组的作用,然后就可以计算每个i结尾的满足题意的串个数. #include<cstdio> #include<cstdlib> #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<memory.h> using namespace std;…

Problem Description It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example: s: "abab" The prefixes are: "…