ACM training II C

KMP is helpless to me, and I am really worried about KMP. It's just an algorithm. I thought it could be done by the stuff around my neck. Now it's okay-it's so big that I have a big head. I want to write a value for next, five words: That's not a problem. When it comes to practical application, I don't know what the next function is and what it can do. It's like seeing two points --

I 've been searching for the same string for a long time, for example, acmacmacmacmacma. Just look at next from the back, for example, the last next [15] = 13, it indicates that the first 13 strings are the same as the last 13 digits, and the total length is 16-13 = 3, which is used as a solution. Next, let's look at next [13] And keep searching for the result, pay attention to the format when outputting the data. I handed in four errors here.

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: 31 2 3Case #2: 63 6 9 12 15 16Case #3: 43 6 9 10Case #4: 29 12

#include <iostream>#include<cstring>using namespace std;int a[1000100],next[1000100];int m;char s[1000100];void getNext(){    int j;    next[0] = 0; next[1] = 0;    for(int i = 1;i < m;i++)    {        j = next[i];        while(j && s[i]!=s[j])        {            j = next[j];        }        next[i+1] = s[i] == s[j]?j+1:0;    }}int main(){    int n,count,where = 1;    cin >> n;    while(n--)    {        cin >> s;        m = strlen(s);               // cout<< m;        count = 0;        int t = m;        getNext();        while(next[m])        {            a[count++] = t - next[m];            m = next[m];        }        cout << "Case #" << where <<": " << count+1 << endl;        where++;        for(int i = 0;i < count ;i++)        {            cout << a[i] << " ";        }        cout <<t << endl;    }    return 0;}