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…

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): 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…

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,…