fzuProblem 1901 Period II

Problem 1901 Period II

Accept: 107 Submit: 297

Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

For each prefix with length P of a given string S,if
S[i]=S[i+P] for i in [0..SIZE(S)-p-1],

then the prefix is a “period” of S. We want to all the periodic prefixs.

Input

Input contains multiple cases.
The first line contains an integer T representing the number of cases. Then following T cases.

Each test case contains a string S (1 <= SIZE(S) <= 1000000),represents the title.S consists of lowercase ,uppercase letter.

Output

For each test case, first output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of periodic prefixs.Then output the lengths of the periodic prefixs in ascending
order.

Sample Input

4oooacmacmacmacmacmafzufzufzufstostootssto

Sample Output

Case #1: 3

1 2 3

Case #2: 6

3 6 9 12 15 16

Case #3: 4

3 6 9 10

Case #4: 2

9 12

#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 1000005;
int next[MAXN],m,cnt;
char pattern[MAXN];
int queue[MAXN],front,rear;

void get_next(){
    int i=0, j=-1;
    next[i]=j;
    while(i<m){
        if(j==-1 || pattern[i]==pattern[j]){
            i++; j++;
            next[i]=j;
        }else j=next[j];
    }
}

int main()
{
    int n;
    cin>>n;
    for(int i=1; i<=n; i++){
        cin>>pattern;
        m=strlen(pattern);
        get_next();
        front=0; rear=0; cnt=1;
        int t=m;
        while(next[t]) {queue[rear++]=next[t]; cnt++; t=next[t];}
        cout<<"Case #"<<i<<": "<<cnt<<endl;
        while(front<rear){
            cout<<m-queue[front++]<<" ";
        }
        cout<<m<<endl;
    }
    return 0;
}