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

一开始写的时候不小心把求ans数组写成了递归，导致overflow。

#include<iostream>
#include<set>
#include<map>
#include<vector>
#include<string>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAXN 1001000
/*
寻找字符串中所有 前缀 和（后缀逆转） 的匹配
将字符串逆转后 添加到原字符串末尾
然后输出所有小于源字符串长度的解
*/
char s[MAXN];
int Next[MAXN],l,cnt;
int ans[MAXN];
void kmp_pre(int m)
{
    int j,k;
    j = 0; k = Next[0] = -1;
    while(j<m)
    {
        if(k==-1||s[j]==s[k])
            Next[++j] = ++k;
        else
            k = Next[k];
    }
}
void Cnt(int p)
{
    while(Next[p]>=0)
    {
        ans[cnt++] = l-Next[p];
        p = Next[p];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int i=1;i<=T;i++)
    {
        cnt = 0;
        scanf("%s",s);
        l = strlen(s);
        kmp_pre(l);
        Cnt(l);
        printf("Case #%d: %d\n",i,cnt);
        for(int j=0;j<cnt;j++)
        {
            if(j) printf(" ");
            printf("%d",ans[j]);
        }
        cout<<endl;
    }
}