# KMP算法讲解

KMP算法是一种改进的字符串匹配算法，由D.E.Knuth，J.H.Morris和V.R.Pratt同时发现，因此人们称它为克努特——莫里斯——普拉特操作（简称KMP算法）。KMP算法的关键是利用匹配失败后的信息，尽量减少模式串与主串的匹配次数以达到快速匹配的目的。具体实现就是实现一个next()函数，函数本身包含了模式串的局部匹配信息。时间复杂度O(m+n)。

Bilibili视频：https://www.bilibili.com/video/av40137935

HDU 1358（Period）

Problem 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 periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.
Input
The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) - the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.
Output
For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
SampleInput
3
aaa
12
aabaabaabaab
0

SampleOutput
Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

 1 #include <bits/stdc++.h>
2 using namespace std;
3
4 int n;
5 string str;
6 int nxt[1000005];
7
8 void getnext(){
9     int i = 0, j = -1, len = str.size();
10     nxt[0] = -1;
11     while(i < len){
12         if(j == -1 || str[i] == str[j])
13             nxt[++i] = ++j;
14         else
15             j = nxt[j];
16     }
17 }
18
19 int main(){
20     ios_base::sync_with_stdio(false),cout.tie(0),cin.tie(0);
21     int tot = 1;
22     while(cin>>n && n){
23         cin>>str;
24         getnext();
25         cout << "Test case #" << tot++ << endl;
26         for(int i = 2; i <= n; i++){
27             if(nxt[i] != 0 && i % (i - nxt[i]) == 0)
28                 cout << i << " " << i/(i - nxt[i]) << endl;
29         }
30         cout << endl;
31     }
32
33     return 0;
34 }

posted @ 2019-01-07 17:23  Xenny  阅读(849)  评论(0编辑  收藏