題目連結:

描述：

給出一字串，求所有公共前後綴長度，從小到大輸出，顯然字串長度也為一個公共前後綴，且為最長的

Description

The little cat is so famous, that many couples tramp over hill and
dale to Byteland, and asked the little cat to give names to their
newly-born babies. They seek the name, and at the same time seek the
fame. In order to escape from such boring job, the innovative little
cat works out an easy but fantastic algorithm:

Step1. Connect the father’s name and the mother’s name, to a new
string S. Step2. Find a proper prefix-suffix string of S (which is not
only the prefix, but also the suffix of S).

Example: Father=’ala’, Mother=’la’, we have S = ‘ala’+’la’ = ‘alala’.
Potential prefix-suffix strings of S are {‘a’, ‘ala’, ‘alala’}. Given
the string S, could you help the little cat to write a program to
calculate the length of possible prefix-suffix strings of S? (He might
thank you by giving your baby a name:)

Input

The input contains a number of test cases. Each test case occupies a
single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <=
Length of S <= 400000.

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby’s name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

思路：

利用next陣列，abcab ……… abcab 最後一個字元的next值為以該字母結束的最長前後綴長度，再到公共綴中找，公共前後綴，必然是整個串的公共前後綴

/*************************************************************************
    > File Name: poj_2752.cpp
    > Author: dulun
    > Mail: [email protected]
    > Created Time: 2016年03月21日 星期一 12時25分35秒
 ************************************************************************/

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define LL long long
using namespace std;

const int N = 400009;
char a[N];
int nxt[N];
int sum[N];

void get_nxt()
{
    int k = 0; 
    int l = strlen(a);
    memset(nxt, 0, sizeof(nxt));
    for(int i = 1; i < l; i++)
    {
        while(k && a[i] != a[k]) k = nxt[k-1];
        if(a[i] == a[k]) k++;
        nxt[i] = k;
    }
}

int main()
{
    while(scanf("%s", a) != EOF)
    {
        get_nxt();
        int l = strlen(a);

        memset(sum, 0, sizeof(sum));      
        int k = 0;
        int m = l;
        while(m)
        {
            sum[k++] = nxt[m-1];
            m = nxt[m-1];
        }
        for(int i = k-2; i>=0; --i) printf("%d ", sum[i]);
        printf("%d\n", l);
    }

    return 0;
}