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　　题目传送门

Seek the Name, Seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23798		Accepted: 12431

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

 

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　　分析：

　　又是一道神奇的$KMP$题，又是需要掌握$next$数组的正确姿势。（咦，我为什么要说又？）

　　首先一个字符串它自己肯定算一个答案。求出$next$数组然后从$next[n]$开始计数，只要当前$next[]$不为$0$，那么就记录新答案为$next[]$值，然后继续。至于正确性可以自己证明，只要深入理解了$next$数组，就不难理解了。

　　（当然，这题可以用$hash$做，且复杂度也差不多是$O(n)$的）

　　Code：

 

//It is made by HolseLee on 11th Aug 2018
//POJ2752
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
using namespace std;

const int N=2e6+7;
int n,nxt[N],k,ans[N],cnt;
char s[N];

int main()
{
    while(scanf("%s",s)!=EOF){
        n=strlen(s);
        nxt[0]=nxt[1]=0;k=0;
        for(int i=1;i<n;++i){
            while(k&&s[i]!=s[k])
            k=nxt[k];
            nxt[i+1]=(s[i]==s[k]?++k:0);
        }
        k=nxt[n];cnt=1;ans[1]=n;
        while(k){
            ans[++cnt]=k;
            k=nxt[k];
        }
        for(int i=cnt;i>=1;--i)
        printf("%d ",ans[i]);
        printf("
");
    }
    return 0;
}