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…

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…

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): 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): 14096    Accepted Submission(s): 6462 Problem Description It is well known that AekdyCoin is good at string problems as well as…

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…

题意: 求给定字符串,包含的其前缀的数量. 分析: 就是求所有前缀在字符串出现的次数的和,可以用KMP的性质,以j结尾的串包含的串的数量,就是next[j]结尾串包含前缀的数量再加上自身是前缀,dp[i]表示以i为结尾包含前缀的数量,则dp[i]=dp[next[i]]+1,最后求和即可. #include <map> #include <set> #include <list> #include <cmath> #include <queue>…

题意:给一个字符串,问该字符串的所有前缀与该字符串的匹配数目总和是多少. 此题要用KMP的next和DP来做. next[i]的含义是当第i个字符失配时,匹配指针应该回溯到的字符位置. 下标从0开始. 设j=next[i],那么 如果j==0,回溯到起点说明该字符不匹配. 其他情况,说明字符串S[0,...j-1]与字符串S[0,..i-1]的某个后缀(准确的说是S[i-j,i-1])相同,这样的话,S[0,..i-1]的后缀(S[i-j,i-1])一定包含字符串S[0,..i-1]的后缀能够匹…

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,…

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…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11607    Accepted Submission(s): 5413Problem Description It is well known that AekdyCoin is good at string problems as well as number theory probl…

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…

[题意概述] 给定一个文本字符串,找出所有的前缀,并把他们在文本字符串中的出现次数相加,再mod10007,输出和. [题目分析] 利用kmp算法的next数组 再加上dp [存在疑惑] 在分析next数组和dp之间的关系,结论是 dp[i] = (dp[next[i]]+1); 搞不懂之间存在的联系 [AC] #include <bits/stdc++.h> ],next[]; ]; void getnext() { ,j=-; next[]=-; while(i<m) { ||s[i…

题意:统计前缀在串中出现的次数 思路:next数组,递推 #include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define MaxSize 200005 #define Mod 10007 char str[MaxSize]; int _next[MaxSize]; int dp[MaxSize]; int len; void GetNext(char t[]){//…

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…

题解:利用next数组来保存前缀位置,递推求解. #include <cstdio> #include <cstring> char pat[200005]; int next[200005],M,f[200005]; const int MOD=10007; int getnext(){ int i=1,j=0;next[1]=0; while(i<M){ if(j==0||pat[j]==pat[i])next[++i]=++j; else j=next[j]; } }…

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…

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…

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…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=3336 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10917    Accepted Submission(s): 5083 Problem Description It is well known that AekdyCoin is good a…

Another Meaning 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5763 Description As is known to all, in many cases, a word has two meanings. Such as "hehe", which not only means "hehe", but also means "excuse me". Today, ?? i…

B. Obsessive String   Hamed has recently found a string t and suddenly became quite fond of it. He spent several days trying to find all occurrences of t in other strings he had. Finally he became tired and started thinking about the following proble…

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…

Infinite monkey theorem Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1702 Accepted Submission(s): 882 Problem Description Could you imaging a monkey writing computer programs? Surely monkeys are…

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…