牛客网多校第3场Esort string （kmp）

链接：https://www.nowcoder.com/acm/contest/141/E
来源：牛客网

题目描述
Eddy likes to play with string which is a sequence of characters. One day, Eddy has played with a string S for a long time and wonders how could make it more enjoyable. Eddy comes up with following procedure:

. For each i in [,|S|-], let Si be the substring of S starting from i-th character to the end followed by the substring of first i characters of S. Index of string starts from .
. Group up all the Si. Si and Sj will be the same group if and only if Si=Sj.
. For each group, let Lj be the list of index i in non-decreasing order of Si in this group.
. Sort all the Lj by lexicographical order.

Eddy can't find any efficient way to compute the final result. As one of his best friend, you come to help him compute the answer!
输入描述:
Input contains only one line consisting of a string S.

≤ |S|≤
S only contains lowercase English letters(i.e. ).
输出描述:
First, output one line containing an integer K indicating the number of lists.
For each following K lines, output each list in lexicographical order.
For each list, output its length followed by the indexes in it separated by a single space.
示例1
输入

复制
abab
输出

复制

示例2
输入

复制
deadbeef
输出

复制

字符串哈希超时，还是太菜看到大佬用unordered map写哈希过了，

kmp 求循环节；

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define ll long long
#define maxn 1000010
char s[maxn];
int len, Next[maxn];
void makeNext(int tlen) {
    int j = , k = -;
    Next[] = -;
    while (j < tlen) {
        if (k == - || s[j] == s[k]) {
            Next[++j] = ++k;
        } else {
            k = Next[k];
        }
    }
}

int main()
{
    cin>>s;
    len=strlen(s);
    makeNext(len);
    int x=len-Next[len];
    if (len % x!= ) {
        printf("%d",len);
        for(int i = ; i < len; i++) {
            printf("1 %d\n",i);
        }
    }
    else {
        printf("%d\n",x);
        for(int i = ; i < x; i++) {
            printf("%d",len/x);
            for(int j = i; j < len; j += x)
                printf(" %d",j);
            printf("\n");
        }
    }

    return ;
}