A Secret Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)Total Submission(s): 830    Accepted Submission(s): 323 Problem Description Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,whi…

题意 : 给出两个字符串,现在需要求一个和sum,考虑第二个字符串的所有后缀,每个后缀对于这个sum的贡献是这个后缀在第一个字符串出现的次数*后缀的长度,最后输出的答案应当是 sum % 1e9+7 分析 : 有两种做法,如果会拓展KMP的话可以说这就是一道模板题了,拓展KMP专门就是找最长公共前缀的长度,首先将给出的两个字符串双双反转,用模式串去跟主串跑一遍拓展KMP,得到 extend 数组,然后只要遍历一遍 extend 数组,每一个 extend[i] 表示模式串里面前 extend[i…

题意:给定两个串,求其中一个串 s 的每个后缀在另一个串 t 中出现的次数. 析:首先先把两个串进行反转,这样后缀就成了前缀.然后求出 s 的失配函数,然后在 t 上跑一遍,如果发现不匹配了或者是已经完全匹配了,要计算,前面出现了多少个串的长度匹配也就是 1 + 2 + 3 + .... + j 在 j 处失配,然后再进行失配处理.注意别忘了,匹配完还要再加上最后的. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000"…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6153 题意:给了串s和t,要求每个t的后缀在在s中的出现次数,然后每个次数乘上对应长度求和. 解法:关键在于想到把s和t都翻转之后,把t求next,然后用t去匹配s,在匹配过程中把fail指针跳到的地方加1,但是还没完,最后需要反向遍历第二个串将大串对小串的贡献加上去就可以了. 这道题是很多现场AC的代码是有漏洞的,比如bazbaba,bazbaba这个答案是34,但是很多现场AC的代码会输出31.…

题目链接 Problem Description Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell: Suffix(S2,i) = S2[i...len].Ni is…

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…

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…

Problem Description Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell: Suffix(S2,i) = S2[i...len].Ni is the…

A Secret Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)Total Submission(s): 461    Accepted Submission(s): 182 Problem Description Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,whi…

题目大意: 给你两个字符串A,B,现在要你求B串的后缀在A串中出现的次数和后缀长度的乘积和为多少. 题解: 扩展KMP模板题,将A和B串都逆序以后就变成了求前缀的问题了,扩展KMP求处从i位置开始的最长公共前缀存于数组. 最后通过将数组的值不为0的进行一个桶计数,倒着进行一下求和就可以了.注意,在这个题目上扩展kmp处理出来的是 ex[ i ]数组是 A串的每个从 i 位置开始的后缀 ,与B串的最长公共前缀长度,那么这样B串在A串上匹配的情况就一目了然了. #include<bits/stdc+…