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…

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…

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…

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…

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"…

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): 6062    Accepted Submission(s): 2810 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…

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…

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…

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…

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[]){//…

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]; } }…

A Secret Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others) Total Submission(s): 2523    Accepted Submission(s): 934 Problem Description Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,w…

链接: https://vjudge.net/problem/HDU-3336 题意: 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"…

今天是字符串填坑的一天,首先填的第一个坑是扩展KMP.总结一下KMP和扩展KMP的区别. 在这里s是主串,t是模式串. KMP可以求出的是以s[i]为结尾的串和 t前缀匹配的最长的长度.假如这个长度是L的话,则: s[i-L+1...i]=t[0...L] 而所谓的失配指针f[i]指的就是当前i点失配时要匹配的长度,实际是用t文本串去匹配t. 扩展KMP则是以s[i]为起始的串和 t前缀匹配的最长的长度. 假如这个长度的话,则: s[i..i+L-1]=t[0...L] 扩展KMP里的nxt数组…

Problem Description Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was preparing for sending it to ykwd. They had agreed that each letter of these messages would be transfered to a…

http://acm.hdu.edu.cn/showproblem.php?pid=4333 题意 一个数字,依次将第一位放到最后一位,问小于本身的数的个数及等于本身的个数和大于本身的个数,但是要注意重复的不再计算 分析 当这个串有循环节时才会出现重复串,用KMP的next数组来计算循环节:len-next[len].最后将得到的答案除以循环节个数即可.然后用扩展KMP,求出主串的后缀与模式串的前缀的最长公共前缀,这里的模式串是原串,而主串则是将两个原串连接起来.利用extend数组,当exte…

A Secret Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)Total Submission(s): 1530    Accepted Submission(s): 570 Problem Description Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,wh…

链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3587 题意:给出两个字符串S和T.S,T<=100000.拿出S的两个子串(能够重叠),将两个子串连接起来成为字符串T的方法有多少种. 思路:用扩展KMP求出S的从每位開始的子串与T的公共前缀,再将两个子串翻转,再用扩展KMP求出S反的从每位開始的子串与T反的公共前缀.找出当中和为T子串长度的S公共前缀和S反的公共前缀的数量,相乘为结果. 代码: #include…

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…

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…