ACM训练二C题

kmp对我真无奈，我对kmp真苦恼。就是个算法嘛，我以为凭我脖子上的东西可以搞定，现在好了--搞得我头都大了。要我写个啥next的值，五个字：那都不是事。一到啥实际应用吧，我意识不行了，搞不清next到底有什么作用，能干什么。就好比见到了二分啊--

此题的意思呢，我弄了很久，其实是找相同串，比如ACMACMACMACMACMA，从后往前看next就行了，比如最后一个next[15] = 13，代表前13个字符串和后13位相同，直接用总长16 - 13 = 3，为一个解，接下来看next[13]了，一直这样找出结果，输出时一定注意格式，我交了4次全错在这里。

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

4
ooo
acmacmacmacmacma
fzufzufzuf
stostootssto

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;

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;
}