题目来源：POJ 3882 Stammering Aliens

题意：给你m一个一个字符串 求至少出现m次的最长字符串 可以在字符串中重叠出现

思路：二分长度l 然后从height数组中找长度大于等于l的前缀

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 40010;
char s[maxn];
int sa[maxn];
int t[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];
int pos, lim;
void build_sa(int m, int n)
{
	int i, *x = t, *y = t2;
	for(i = 0; i < m; i++)
		c[i] = 0;
	for(i = 0; i < n; i++)
		c[x[i] = s[i]]++;
	for(i = 1; i < m; i++)
		c[i] += c[i-1];
	for(i = n-1; i >= 0; i--)
		sa[--c[x[i]]] = i;
	for(int k = 1; k <= n; k <<= 1)
	{
		int p = 0;
		for(i = n-k; i < n; i++)
			y[p++] = i;
		for(i = 0; i < n; i++)
			if(sa[i] >= k)
				y[p++] = sa[i] - k;
		for(i = 0; i < m; i++)
			c[i] = 0;
		for(i = 0; i < n; i++)
			c[x[y[i]]]++;
		for(i = 0; i < m; i++)
			c[i]+= c[i-1];
		for(i = n-1; i >= 0; i--)
			sa[--c[x[y[i]]]] = y[i];
		swap(x,y);
		p = 1; x[sa[0]] = 0;
		for(i = 1; i < n; i++)
			x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
		if(p >= n)
			break;
		m = p;
	}
}


void getHeight(int n)
{
	int k = 0;
	for(int i = 0; i <= n; i++)
		rank[sa[i]] = i;
	for(int i = 0; i < n; i++)
	{
		if(k)
			k--;
		int j = sa[rank[i]-1];
		while(s[i+k] == s[j+k])
			k++;
		height[rank[i]] = k;
	}
}

bool ok(int l, int n)
{
	int flag = 0;
	int cnt = 1;
	pos = 0;
	int p = -1;
	for(int i = 1; i <= n; i++)
	{
		if(height[i] >= l)
		{
			cnt++;
			p = max(p, sa[i]);
		}
		else
		{
			
			p = sa[i];
			cnt = 1;
		}
		
		if(cnt >= lim)
		{
			flag = 1;
			pos = max(pos, p);
		} 
	}
	if(flag)
		return true;
	return false;
}
int main()
{
	int l, r, n;
	while(scanf("%d", &lim) && lim)
	{
		scanf("%s", s);
		n = strlen(s);
		if(lim == 1)
		{
			printf("%d 0\n", n);
			continue;
		}
		l = 1;
		r = n;
		int ans = -1;
		int ans2;
		build_sa(256, n+1);
		getHeight(n);
	
		while(l <= r)
		{
			int m = (l + r) >> 1;
			
			if(ok(m, n))
			{
				ans = m;
				ans2 = pos;
				l = m+1;
			}
			else
				r = m-1;
		}

		if(ans == -1)
			puts("none");
		else
		{
			printf("%d %d\n", ans, ans2);
		}
		
	}
	return 0;
	
}