读不懂的流年

在KMP算法中，求解next[]数组是解题的关键，下面我来总结一下我求解的代码： void getNext(int *next) {     int i,j;     i=0;//i指代后缀     j=-1;//j指代前缀     next[0]=-1;     while(i     {         if(j==-1||p[i]==p[j])         {

Problem Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b

题目链接 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 = "abcdef". If we think of concatenation as multipl

Description There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new s

题目链接 Problem Description 一块花布条，里面有些图案，另有一块直接可用的小饰条，里面也有一些图案。对于给定的花布条和小饰条，计算一下能从花布条中尽可能剪出几块小饰条来呢？     Input 输入中含有一些数据，分别是成对出现的花布条和小饰条，其布条都是用可见ASCII字符表示的，可见的ASCII字符有多少个，布条的花纹也有多少种花样。

题目链接 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

第一部分 1、KMP 算法思想     普通的字符串匹配算法必须要回溯。但回溯就影响了效率，回溯是由T串本身的性质决定的，是因为T串本身有前后'部分匹配'的性质。像上面所说如果主串为abcdef这样的，大没有回溯的必要。     改进的地方也就是这里，我们从T串本身出发，事先就找准了T自身前后部分匹配的位置，那就可以改进算法。     如果不用回溯，那模式串下一个位置从哪里开始呢？   

题目链接 Description For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodi