假设求串′ababaaababaa′的next数组 模式串 a b a b a a a b a b a a 下标 1 2 3 4 5 6 7 8 9 10 11 12 1.前两位:next数组前两位一定是0,1 即前两位ab对应的next数组为01,则: 模式串 a b a b a a a b a b a a 下标 1 2 3 4 5 6 7 8 9 10 11 12 next数组 0 1 2.接下来看第三位,按照next数组求解方法.第三位a的前一位为第
Period Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2398 Accepted Submission(s): 1187 Problem Description For each prefix of a given string S with N characters (each character has an ASCII
之前在学KMP算法时一直理解不了获取next数组的函数是如何实现的,现在大概知道怎么一回事了,记录一下我对获取next数组的理解. KMP算法实现的原理就不再赘述了,先上KMP代码: 1 void getNext(char *pat, int *next) { 2 next[0] = -1; 3 int m = strlen(pat); 4 for (int i = 1; i < m; i++) { 5 int j = next[i - 1]; 6 while (j != -1 &&
查看原题 题意大致是:给你一个字符串算这里面全部前缀出现的次数和.比方字符串abab,a出现2次.ab出现2次,aba出现1次.abab出现1次.总计6次. 而且结果太大.要求对1007进行模运算. AC代码 #include <iostream> using namespace std; #include <string> string s; int n,Next[200005]; void getNext() { int len = n; Next[0]=-1; int i=0
Power Strings Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 30069 Accepted: 12553 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "