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: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…

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&…

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…

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…

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"…

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…

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…

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): 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): 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): 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…

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…

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…

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: "…

一道字符串匹配的题目,仅仅借此题练习一下KMP 因为这道题目就是要求用从头开始的n个字符串去匹配原来的字符串,很明显与KMP中求next的过程很相似,所以只要把能够从头开始匹配一定个数的字符串的个数加起来就OK了(再此结果上还应该加上字符串的长度,因为每个从头开始的字符串本身也可以去匹配自己的),即将next中值不为-1和0的个数统计出来即可. 用GCC编译的,时间用了46MS. #include <stdio.h> #include <string.h> #define MAXL…

参考连接: KMP+DP: http://www.cnblogs.com/yuelingzhi/archive/2011/08/03/2126346.html 另外给出一个没用dp做的:http://blog.sina.com.cn/s/blog_82061db90100usxw.html 题意: 给出一个字符串,求它的各个前缀在字符串中出现的次数总和. 思路:记 dp[i] 为前 i 个字符组成的前缀出现的次数则 dp[next[i]]+=dp[i] dp[i]表示长度为i的前缀出现的次数,初…

题目 以下不是KMP算法—— 以下是kiki告诉我的方法,好厉害的思维—— 就是巧用标记,先标记第一个出现的所有位置,然后一遍遍从标记的位置往下找. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { ],shunxu; ]; scanf("%d",&t); while(t--) { memset(xiabiao,…