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：求所有的循环节
思路：循环ans=ne[ans],len-ans

#include <iostream>
#include<string.h>
using namespace std;
const int maxn = 1e6+10;
char a[maxn];
int ne[maxn],sum[maxn];
int main()
{
    int t;
    int num=0;
    scanf("%d",&t);
    while(t--)
    {
        memset(ne,0,sizeof(ne));
        scanf("%s",a+1);
        int len=strlen(a+1);
        for(int i=2,j=0;i<=len;i++)
        {
            while(j&&a[i]!=a[j+1]) j=ne[j];
            if(a[i]==a[j+1]) j++;
            ne[i]=j;
        }
        int ans=len,cnt=0;
        while(ans)
        {
            ans=ne[ans];
            sum[++cnt]=len-ans;
        }
        printf("Case #%d: %d\n",++num,cnt);
        for(int i=1;i<=cnt-1;i++)
            printf("%d ",sum[i]);
        printf("%d\n",sum[cnt]);
    }
    return 0;
}