Seek the Name, Seek the Fame （POJ-2752）

Seek the Name, Seek the Fame （POJ-2752）

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

题意：找出所给串的所有前缀长度，使得所给串中这个长度的前缀==这个长度的后缀 
思路：s是既是自身的前缀也是自身的后缀。这个问题利用KMP算法的next[]数组来解。首先对于输入的字符串s，计算其next[]数组。然后，根据next[]数组的值进行进一步的计算。还需要知道的是next[]数组的定义。对于字符串s的第i个字符s[i]，next[i]定义为字符s[i]前面最多有多少个连续的字符和字符串s从初始位置开始的字符匹配。从后到前匹配前缀和后缀。如果前缀与后缀匹配，则下一个前缀与后缀匹配的字符串只能在前缀中。

AC代码：

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <string>
#include <string.h>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#define PI 3.1415926
#define maxx 1000010
using namespace std;
char s[maxx];
int nex[maxx];
int ans[maxx];
void Getnextval(char *p)
{
    int plen=strlen(p);
    int i=0;
    int j=-1;
    nex[0]=-1;
    while (i<plen)
        if(j==-1 || p[i]==p[j])
            nex[++i]=++j;
        else
            j=nex[j];
}
int main()
{
    int i,j,k;
    while(~scanf("%s",s))
    {
        k=0;
        Getnextval(s);
        int slen=strlen(s);
        j=nex[slen];
        while(j>0)
        {
            ans[k++]=j;
            j=nex[j];
        }
        for(i=k-1; i>=0; i--)
            printf("%d ",ans[i]);
        printf("%d\n",slen);
    }
    return 0;
}