记录编号	380812	评测结果	AAAAAAAAAAAAAAA
题目名称	乐曲主题	最终得分	100
用户昵称	可以的.	是否通过	通过
代码语言	C++	运行时间	0.024 s
提交时间	2017-03-10 11:35:45	内存使用	0.89 MiB

#include <stdio.h>
#include <ctime>
#include <algorithm>
#include <cstring>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <cstdlib>
#include <iostream>
#include <bitset>
#include <cmath>
#include <vector>
#define Mem(a, b) memset(a, b, sizeof a)
#define Mcpy(a, b) memcpy(a, b, sizeof a)
#define RG register
#define IL inline
using namespace std;
typedef long long LL;
typedef unsigned int uint;
typedef unsigned long long ull;
IL void qkin(RG int &res){
	RG int x,f=1; RG char ch;
	while(ch=getchar(),ch<'0'||ch>'9')if(ch=='-')f=-1;x=ch-48;
	while(ch=getchar(),ch>='0'&&ch<='9')x=x*10+ch-48;res=x*f;}
IL void read(RG int &A){ qkin(A); }
IL void read(RG int &A,RG int &B){ qkin(A),qkin(B); }
IL void read(RG int &A,RG int &B,RG int &C){ qkin(A),qkin(B),qkin(C); }
#define Gets(s) scanf("%s", s+1);
const double INF = 1e60;
const int inf = 0x7f7f7f7f;
const double Pi = acos(-1);
const double eps = 1e-8;
const int maxn = 5010*2;

int n,w[maxn],cf[maxn],rt,last,cnt;
struct SAM
{
	int root,val; 
	map<int, int> next;
}a[maxn];
void add(int c){
	int p = last, np = ++cnt;
	a[np].val = a[p].val + 1;
	while(p && !a[p].next[c]){
		a[p].next[c] = np;
		p = a[p].root;
	}
	if(!p) a[np].root = rt;
	else {
		int q = a[p].next[c];
		if(a[q].val == a[p].val+1) a[np].root = q;
		else {
			int nq = ++cnt;
			a[nq] = a[q];
			a[nq].val = a[p].val+1;
			a[np].root = a[q].root = nq;
			while(p && a[p].next[c] == q){
				a[p].next[c] = nq;
				p = a[p].root;
			}
		}
	}
	last = np;
}
int bkt[maxn],d[maxn];
int size[maxn];
int L[maxn],R[maxn];
int main(){
#define HAHA LALA
#ifdef HAHA
	freopen("theme.in", "r", stdin);
	freopen("theme.out", "w", stdout);
#endif
	rt = last = ++cnt;
	read(n); for(int i=1; i<=n; i++) read(w[i]);
	for(int i=1; i<n; i++) {
		cf[i] = w[i+1] - w[i];
		add(cf[i]); 
	}	
	int p = rt;
	Mem(L, 63); Mem(R, 0);
	for(int i=1; i<n; i++) {
		size[p = a[p].next[cf[i]]] = 1;
		L[p] = R[p] = i;
	}
	for(int i=1; i<=cnt; i++) bkt[a[i].val]++;
	for(int i=1; i<=cnt; i++) bkt[i] += bkt[i-1];
	for(int i=cnt; i>=1; i--) d[bkt[a[i].val]--] = i;
	for(int i=cnt; i>=1; i--){
		int x = d[i], fa = a[x].root;
		L[fa] = min(L[fa], L[x]);
		R[fa] = max(R[fa], R[x]);
		size[fa] += size[x];
	}
	int ans = 0;
	for(int i=1; i<=cnt; i++){
		if(size[i] >= 2){
			ans = max(ans, min(R[i]-L[i], a[i].val+1));
		}
	}
	ans = ans>=5?ans:0;
	printf("%d\n", ans);
	return 0;
}