# poj2752:KMP的简单应用

Seek the Name, Seek the Fame
 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 16388 Accepted: 8330

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

KMP next(fail)数组的简单应用，从next[len]一直递推到底，输出长度即可。
next数组递推过程中一定是严格递减的，所以我们可以不用排序，直接逆序输出即可。

``````/*Source Code
Problem: 2752		User: aclolicon
Memory: 3188K		Time: 110MS
Language: G++		Result: Accepted
Source Code*/
#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXL 400050
using namespace std;
int fail[MAXL], len, ans, tot;
int a[MAXL];
char c[MAXL];
void getfail(){
int i = 0, j = -1;
fail[0] = -1;
while(i != len){
if (j == -1 || c[i] == c[j]) fail[++i] = ++j;
else j = fail[j];
}
}
``````int main(){
while(~scanf("%s", c)){
ans = 0, tot = 0;
len = strlen(c);
getfail();
ans = fail[len];
a[tot++] = len; //整个字符串肯定是
while(ans > 0){
a[tot++] = ans;
ans = fail[ans];
}
for (int i = tot - 1; i >= 0; i--) cout << a[i] << (i == 0? '\n': ' ');
}
}``````