SPOJ REPEATS Repeats

A string s is called an (k,l)-repeat if s is obtained by concatenating k>=1 times some seed string t with length l>=1. For example, the string

s = abaabaabaaba

is a (4,3)-repeat with t = aba as its seed string. That is, the seed string t is 3 characters long, and the whole string s is obtained by repeating t 4 times.

Write a program for the following task: Your program is given a long string u consisting of characters ‘a’ and/or ‘b’ as input. Your program must find some (k,l)-repeat that occurs as substring within u with k as large as possible. For example, the input string

u = babbabaabaabaabab

contains the underlined (4,3)-repeat s starting at position 5. Since u contains no other contiguous substring with more than 4 repeats, your program must output the maximum k.

Input

In the first line of the input contains H- the number of test cases (H <= 20). H test cases follow. First line of each test cases is n - length of the input string (n <= 50000), The next n lines contain the input string, one character (either ‘a’ or ‘b’) per line, in order.

Output

For each test cases, you should write exactly one interger k in a line - the repeat count that is maximized.

Example

Input:
1
17
b
a
b
b
a
b
a
a
b
a
a
b
a
a
b
a
b

Output:
4

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define fi first
#define se second
#define mp(i,j) make_pair(i,j)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
const int read()
{
	char ch = getchar();
	while (ch<'0' || ch>'9') ch = getchar();
	int x = ch - '0';
	while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';
	return x;
}
int T = 0;

struct Sa
{
	char s[N];
	int rk[2][N], sa[N], h[N], w[N], now, n;
	int rmq[N][20], lg[N];

	bool GetS()
	{
		scanf("%d", &n);
		rep(i, 1, n) scanf("%s", s + i);
		return true;
	}

	void getsa(int z, int &m)
	{
		int x = now, y = now ^= 1;
		rep(i, 1, z) rk[y][i] = n - i + 1;
		for (int i = 1, j = z; i <= n; i++)
			if (sa[i] > z) rk[y][++j] = sa[i] - z;

		rep(i, 1, m) w[i] = 0;
		rep(i, 1, n) w[rk[x][rk[y][i]]]++;
		rep(i, 1, m) w[i] += w[i - 1];
		per(i, n, 1) sa[w[rk[x][rk[y][i]]]--] = rk[y][i];
		for (int i = m = 1; i <= n; i++)
		{
			int *a = rk[x] + sa[i], *b = rk[x] + sa[i - 1];
			rk[y][sa[i]] = *a == *b&&*(a + z) == *(b + z) ? m - 1 : m++;
		}
	}

	void getsa(int m)
	{
		//n = strlen(s + 1);
		rk[1][0] = now = sa[0] = s[0] = 0;
		rep(i, 1, m) w[i] = 0;
		rep(i, 1, n) w[s[i]]++;
		rep(i, 1, m) rk[1][i] = rk[1][i - 1] + (bool)w[i];
		rep(i, 1, m) w[i] += w[i - 1];
		rep(i, 1, n) rk[0][i] = rk[1][s[i]];
		rep(i, 1, n) sa[w[s[i]]--] = i;

		rk[1][n + 1] = rk[0][n + 1] = 0;	//多组的时候容易出bug
		for (int x = 1, y = rk[1][m]; x <= n && y <= n; x <<= 1) getsa(x, y);
		for (int i = 1, j = 0; i <= n; h[rk[now][i++]] = j ? j-- : j)
		{
			if (rk[now][i] == 1) continue;
			int k = n - max(sa[rk[now][i] - 1], i);
			while (j <= k && s[sa[rk[now][i] - 1] + j] == s[i + j]) ++j;
		}
	}

	void getrmq()
	{
		h[n + 1] = h[1] = lg[1] = 0;
		rep(i, 2, n) rmq[i][0] = h[i], lg[i] = lg[i >> 1] + 1;
		for (int i = 1; (1 << i) <= n; i++)
		{
			rep(j, 2, n)
			{
				if (j + (1 << i) > n + 1) break;
				rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);
			}
		}
	}

	int lcp(int x, int y)
	{
		int l = min(rk[now][x], rk[now][y]) + 1, r = max(rk[now][x], rk[now][y]);
		return min(rmq[l][lg[r - l + 1]], rmq[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);
	}

	void work()
	{
		GetS();	getsa(300);	getrmq();
		int ans = 1;
		rep(L, 1, n)
		{
			for (int i = 1; i + L <= n; i += L)
			{
				int R = lcp(i, i + L);
				ans = max(ans, R / L + 1);
				if (i >= L - R % L)
				{
					ans = max(lcp(i - L + R%L, i + R%L) / L + 1, ans);
				}
			}
		}
		printf("%dn", ans);
	}
}sa;

int main()
{
	T = read();
	while (T--) sa.work();
	return 0;
}