PKU 2752 seek the Name and seek the Fame (KMP)

Seek the Name and seek the Fame

Time Limit: 2000MS		Memory Limit: 65536K
Total submissions: 9783		accepted: 4693

Description the little cat is so famous, which 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 in the same time seek the fame. In order to escape from such boring job, the innovative little cat works out a easy but fantastic algorithm: 
br> 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 isn't 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 str Ings of S? (He might thank your by giving your baby a name:)  

Input the input contains a number of test cases. Each test case is occupies a single and 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 an integer numbers in increasing order, denoting the possible length of The new baby ' s name.

Sample Input

Ababcababababcabab
AAAAA

Sample Output

2 4 9 1 2 3 4 5

Source POJ Monthly--2006.01.22,zeyuan Zhu
To find the length of the substring that is both the prefix of the string and the suffix of the string: Use the KMP value flexibly and save the data with the stack. The value of next is known to represent the next "I"-1 characters and the prefix of the next "I"-1 characters in front of the position I, and the equal pass of the iteration proves that each I=next "i+1"-1 is a substring that is both a prefix and a suffix

#include <stdio.h>
#include <string.h>
char s[400005];
int next[400005],sta[400005],all;
void Get_next ()
{
    int len=strlen (s+1), i=1,j;

    next[1]=j=0;
    while (I<=len)
    {
        if (j==0| | S[I]==S[J]) next[++i]=++j;
        else j=next[j];
    }
void Solve ()
{
    int len=strlen (s+1), I;

    All=0,sta[all++]=len;
    for (i=len;i>1;)
    {
        if (next[i+1]>1) sta[all++]=next[i+1]-1;
        I=next[i+1]-1
    }
}
void print ()
{
    int i;
    for (i=all-1;i>0;i--) printf ("%d", Sta[i]);
    printf ("%d\n", Sta[i]);
int main ()
{
    while (scanf ("%s", &s[1]) >0)
    {
        get_next ();
        Solve ();
        Print ();
    }
    return 0;
}