Seek the Name, Seek the Fame


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Seek the Name, Seek the Fame


Time Limit:

Memory Limit:

Total Submissions:

Accepted:

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

Source

​​POJ Monthly–2006.01.22​​,Zeyuan Zhu


找公共前后缀的最大长度
比如第一个样列，
ababcababababcabab a b a b c a b a b a b a b c a b a b
公共前后缀有：

  • ababcababababcabab a b a b c a b a b a b a b c a b a b
  • ababcabab a b a b c a b a b
  • abab a b a b
  • ab a b

长度分别是 18 9 4 2
next数组记录的其实就是最长公共前后缀的长度

next[i]：字符串 0 到 i 的最长公共前后缀的长度

比如这个样列：
ababcababababcabab a b a b c a b a b a b a b c a b a b

i

next [ i ]

0

next [ 0 ] = -1

1

next [ 1 ] = 0

2

next [ 2 ] = 0

3

next [ 3 ] = 1

4

next [ 4 ] = 2

5

next [ 5 ] = 0

6

next [ 6 ] = 1

7

next [ 7 ] = 2

8

next [ 8 ] = 3

9

next [ 9 ] = 4

10

next [ 10 ] = 3

11

next [ 11 ] = 4

12

next [ 12 ] = 3

13

next [ 13 ] = 4

14

next [ 14 ] = 5

15

next [ 15 ] = 6

16

next [ 16 ] = 7

17

next [ 17 ] = 8

18

next [ 18 ] = 9

得到 next 数组之后，我们从 i=18 开始，
i=18记录 18next[18]=9 i = 18 记 录   18 n e x t [ 18 ] = 9
i=9记录 9next[9]=4 i = 9 记 录   9 n e x t [ 9 ] = 4
i=4记录 4next[4]=2 i = 4 记 录   4 n e x t [ 4 ] = 2
i=2记录 2next[2]=0 i = 2 记 录   2 n e x t [ 2 ] = 0
i=0结束 i = 0 结 束

18,9,4,2 18 , 9 , 4 , 2

答案就是2 4 9 18 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<iostream>
using namespace std;
const int MAXN=400000+7;
char s[MAXN];
int slen,nex[MAXN];

void GetNext()
{
    int i,j;
    i=0;
    j=nex[0]=-1;
    while(i<slen)
    {
        while(-1!=j && s[i] != s[j])    j=nex[j];
        nex[++i]=++j;
    }
}
int main()
{
    while(~scanf("%s", s))
    {
        slen=strlen(s);
        GetNext();
        stack<int> sta;
//        for(int i=0;i<=slen;i++)
//            printf("next[%2d]:%2d\n",i,nex[i]);
        for(int i=slen;~i;i=nex[i])
            if(i>0)
            {
//                printf("next[%2d]:%2d\n",i,nex[i]);
                sta.push(i);
            }
        while(sta.size())
        {
            printf("%d ",sta.top());
            sta.pop();
        }
        printf("\n");
    }
    return 0;
}