kmp（最長前綴與后綴）

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

 

題意：一個字符串，問輸出所有：長度為多少（小於等於總長），能夠由最少2個相同的子字符串組成

 注意：單純是求最長前綴與后綴的題目，最好不要進行next數組優化（匹配題可以使用）

 

#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\n" , ++ans);

        memset(next , 0 , sizeof(next));
        getnext(a , n, next);//求next數組
        for(int i = 2 ; i <= n ; i++)//遍歷每個大於2的字符串
        {
            //i % i (i - next[i])表示該字符串都由若干個（大於1個）相同的字符串組成
            if(next[i] >= 1 && i % (i - next[i]) == 0)
            {
                int l = i - next[i] ;
                printf("%d %d\n" ,i , i / l); // i / l 求由多少個相同的個相同字符串組成
            }
        }
        printf("\n");
    }

    return 0 ;
}