kmp（最长前缀与后缀）

http://acm.hdu.edu.cn/showproblem.php?pid=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.

 

Sample Input

3 aaa 12 aabaabaabaab 0

 

Sample Output

Test case #1

2 2

3 3

Test case #2

2 2

6 2

9 3

12 4

 

Recommend

JGShining

 

题意：一个字符串，问输出所有：长度为多少（小于等于总长），能够由最少2个相同的子字符串组成

 注意：单纯是求最长前缀与后缀的题目，最好不要进行next数组优化（匹配题可以使用）

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
char a[1000009];


void getnext(char *a , int len , int *next)
{
    next[0] = -1 ;
    int k = -1 , j = 0;
    while(j < len)
    {
        if(k == -1 || a[j] == a[k])
        {
            k++;
            j++;
            next[j] = k ;
        }
        else
        {
            k = next[k];
        }
    }
}

int main()
{
    int n , ans = 0 ;
    while(~scanf("%d" , &n) && n)
    {
        int next[1000009];
        scanf("%s" , a );
        printf("Test case #%d
" , ++ans);

        memset(next , 0 , sizeof(next));
        getnext(a , n, next);//求next数组
        for(int i = 2 ; i <= n ; i++)//遍历每个大于2的字符串
        {
            //i % i (i - next[i])表示该字符串都由若干个（大于1个）相同的字符串组成
            if(next[i] >= 1 && i % (i - next[i]) == 0)
            {
                int l = i - next[i] ;
                printf("%d %d
" ,i , i / l); // i / l 求由多少个相同的个相同字符串组成
            }
        }
        printf("
");
    }

    return 0 ;
}