poj 2752 Seek the Name, Seek the Fame

Seek the Name, Seek the Fame

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14092 Accepted: 7009

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

ababcababababcababaaaaa

Sample Output

2 4 9 181 2 3 4 5嗯，题目大一就是说，求字符串中也是后缀的前缀的长度，从小到大排列。

首先它本身就是，然后听说next[lens]记录的就是除去它本身的最长的也是后缀的前缀，然后以此时的状态继续如此，就是递归。我是先把字符串的next数组求出来之前输出然后和测试数据比对的，汗，不是什么好方法。哦，这里只能用next数组，nextval数组会丢失一些数据，渣渣我也不懂。。。

#include<cstdio>#include<cstring>#define len 1000000+10char str[len];int lens,next[len],rec[len];void getnext(){int i=0,j=-1;next[i]=j;while(i<lens){if(j==-1||str[i]==str[j]){i++;j++;next[i]=j;}else j=next[j];}}int main(){while(~scanf("%s",str)){lens=strlen(str);getnext();rec[0]=lens;int i=1;while(next[lens]!=0){rec[i++]=next[lens];lens=next[lens];}for(int j=i-1;j>0;j--)printf("%d ",rec[j]);printf("%d\n",rec[0]);}return 0;}