Q - Period II

Q - Period II

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

题意：

          求所有满足s[i]=s[i+p]的前缀的长度p；

思路:

         ls-next[xun];

 题：

        H - Seek the Name, Seek the Fame

代码：

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MM=1000005;
#define mem(a,b) memset(a,b,sizeof(a))

int ne[MM],a[MM];
char mo[MM];
int lm;
void Get_next()
{
    int i=0,j=-1;
    ne[0]=-1;
    while(i<lm)
    {
        while(j!=-1&&mo[i]!=mo[j])
            j=ne[j];
        ne[++i]=++j;
    }
}
int main()
{
    int ca,cas=1;
    scanf("%d",&ca);
    while(ca--)
    {

        scanf("%s",mo);
        int k=0;
        lm=strlen(mo);
        a[k++]=lm;
        mem(ne,0);
        Get_next();
        int tmp=ne[lm];
        while(tmp)
        {
            a[k++]=lm-tmp;
            tmp=ne[tmp];
        }
        printf("Case #%d: %d\n",cas++,k);
        for(int i=1;i<k;i++)
            printf("%d ",a[i]);
        printf("%d\n",a[0]);
    }
    return 0;
}