poj 2752 找到所有的公共前后缀

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Seek the Name, Seek the Fame
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 19251 Accepted: 9898

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
举个例子  abcabcabc nxet[9]的值为6 
next[6]的值为3 sh说明前六个里前三个和后三个匹配 因为前六个和后六个匹配 所以前三个 和最后三个也匹配了!!!!!!11
next[3]的值为0 所以不停向下找就能找到所有的公共前后缀 
#include <iostream>#include<cstdio>#include<cstring>using namespace std;char a[1000005];int n[1000005];void kmp(int len){    n[0]=-1;    int k=-1;    int j=0;    while(j<=len)    {        if(k==-1||a[j]==a[k])        {            k++;            j++;            n[j]=k;        }        else k=n[k];    }}void print(int len){    if(n[len]>0)   //递归打印 先打了最小的    {        print(n[len]);        printf("%d ",n[len]);    }}int main(){    while(scanf("%s",a)!=-1)    {        int len=strlen(a);        kmp(len);        print(len);        printf("%d\n",len);    }    return 0;}
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