5. 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:
1. For each i in [0,|S|-1], let S i 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 0.
2. Group up all the S i . S i and S j will be the same group if and only if S i =S j .
3. For each group, let L j be the list of index i in non-decreasing order of S i in this group.
4. Sort all the L j 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.
1≤ |S|≤ 10 6
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
abab
输出
2
2 0 2
2 1 3
示例2:
输入
deadbeef
输出
8
1 0
1 1
1 2
1 3
1 4
1 5
1 6
1 7
题目手动复制，可能有点不一样
题解：这道题说把把前i个移到尾 在分个组 其实就是无非找个循环节，输出循环节长度，总长度除以循环节长度,从一个位置开始不断加循环节长度，我一开始直接把全部东西塞到set里求循环节长度，爆内存过55% 然后用kmp求循环节长度，结果过95%然后就蒙逼了，无可奈何最后加了个输入输出挂 结果就过了 都这么骚的吗！！！
#include<bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
typedef long long ll;
#define rep(i, a, n) for (int i=a;i<n;i++)
#define per(i, a, n) for (int i=n-1;i>=a;i--)
#define clr(x) memset(x,0,sizeof(x))
const int INF = 0x3f3f3f3f;
char b[1000005];
int Next[1000005];
int n;
namespace io {
const int L = (1 << 21) + 1;
char ibuf[L], *iS, *iT, obuf[L], *oS = obuf, *oT = obuf + L - 1, c, st[55];
int f, tp;
#define gc() (iS==iT?(iT=(iS=ibuf)+fread(ibuf,1,L,stdin),(iS==iT?EOF:*iS++)):*iS++)
inline void flush() {
fwrite(obuf, 1, oS - obuf, stdout);
oS = obuf;
}
inline void putc(char x) {
*oS++ = x;
if (oS == oT) flush();
}
template<class I>
inline void gi(I &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15);
x *= f;
}
template<class I>
inline void print(I x) {
if (!x) putc('0');
if (x < 0) putc('-'), x = -x;
while (x) st[++tp] = x % 10 + '0', x /= 10;
while (tp) putc(st[tp--]);
}
inline void gs(char *s, int &l) {
for (c = gc(); c < 'a' || c > 'z'; c = gc());
for (l = 0; c <= 'z' && c >= 'a'; c = gc()) s[l++] = c;
s[l] = 0;
}
inline void ps(const char *s) { for (int i = 0, n = strlen(s); i < n; i++) putc(s[i]); }
struct IOFLUSHER {
~IOFLUSHER() { flush(); }
} _ioflusher_;
};
void SetNext()//子串的next数组，如果只看代码的话，很难得出值的确定方法。
{
int i = 0, j = -1;
Next[0] = -1;
while (i < n) {
if (j == -1 || b[i] == b[j])//b为模式数组
{
i++;
j++;
Next[i] = j;
} else
j = Next[j];
}
}
int main() {
// 本地测试
#ifdef ONLINE_JUDGE
#else
freopen("C://Users//yan//Desktop//in.txt","r",stdin);
// freopen("C://Users//yan//Desktop//out.txt","w",stdout);
#endif
ios::sync_with_stdio(false);//取消同步
std::cin.tie(0);//解除cin与cout的绑定，进一步加快执行效率。
io::gs(b, n);
SetNext();
int sum = n - Next[n];
int ssum = n / sum;
io::print(sum);
io::putc('\n');
// cout << sum << endl;
for (int i = 0; i < sum; i++) {
io::print(ssum);
// cout << ssum;
for (int j = i; j < n; j = j + sum) {
io::putc(' ');
io::print(j);
// cout << " " << j;
}
// cout << endl;
io::putc('\n');
}
return 0;
}