Codeforces Round #283 (Div. 2) D. Tennis Game_html/css_WEB-ITnose

The standard schedule seems to be two points?

I preprocessed and marked it, and then enumerated t from 1 to n, because for each t, ​​if s exists, then s is unique, so just enumerate t.

Determine whether t is legal. If you use direct violence, it will time out

I don’t know if the general idea of ​​​​the standard process is the same as mine. Maybe the standard process is not marked but divided into two parts to find

However, my time complexity is better, the approach of n*(1 1/2 1/3...1/n)=n*lnn

#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;struct Ans{	int s,t;	Ans(){}	Ans(int a,int b){s=a;t=b;}	bool operator <(Ans x)const	{		return s!=x.s?s<x.s:t<x.t;	}};vector<Ans>ans;int in[100010];int as[100010];int ap[200010];int bs[100010];int bp[200010];int n;void maketable(){	int i,x,y;	x=y=0;	memset(ap,-1,sizeof(ap));	memset(bp,-1,sizeof(bp));	for(i=0;i<n;i++)	{		if(in[i]==1)			ap[++x]=i;		else			bp[++y]=i;		as[i]=x;		bs[i]=y;	}}int isok(int s){	int t1,t2,x,y;	t1=t2=s;	x=y=0;	while(1)	{		if(ap[t1]==-1&&bp[t2]==-1)			return 0;		if(ap[t1]==n-1&&bp[t2]==-1)		{			++x;			return x<=y?0:x;		}		if(ap[t1]==-1&&bp[t2]==n-1)		{			++y;			return y<=x?0:y;		}		if(bp[t2]==-1||(ap[t1]!=-1&&ap[t1]<bp[t2]))		{			t2=bs[ap[t1]]+s;			t1+=s;			x++;		}		else if(ap[t1]==-1||(bp[t2]!=-1&&ap[t1]>bp[t2]))		{			t1=as[bp[t2]]+s;			t2+=s;			y++;		}		else			return 0;	}}int main(){	int i,t,m;	scanf("%d",&n);	for(i=0;i<n;i++)		scanf("%d",in+i);	maketable();	for(i=1;i<=n;i++)	{		t=isok(i);		if(t!=0)			ans.push_back(Ans(t,i));	}	sort(ans.begin(),ans.end());	m=ans.size();	printf("%d\n",m);	for(i=0;i<m;i++)		printf("%d %d\n",ans[i].s,ans[i].t);}

time limit per test

memory limit per test

input

output

Petya and Gena love playing table tennis. A single match is played according to the following rules: a match consists of multiple sets, each set consists of multiple serves. Each serve is won by one of the players, this player scores one point. As soon as one of the players scores t points, he wins the set; then the next set starts and scores of both players are being set to 0. As soon as one of the players wins the total ofs sets, he wins the match and the match is over. Heres and t are some positive integer numbers.

To spice it up, Petya and Gena choose new numbers s andt before every match. Besides, for the sake of history they keep a record of each match: that is, for each serve they write down the winner. Serve winners are recorded in the chronological order. In a record the set is over as soon as one of the players scores t points and the match is over as soon as one of the players wins sets.

Petya and Gena have found a record of an old match. Unfortunately, the sequence of serves in the record isn't divided into sets and numberss and t for the given match are also lost. The players now wonder what values ​​ofs and t might be . Can you determine all the possible options?

Input

The first line contains a single integer n ? the length of the sequence of games (1?≤?n ?≤?105).

The second line contains n space-separated integersai. Ifai?=?1, then thei-th serve was won by Petya, if ai?=?2, then the i-th serve was won by Gena.

It is not guaranteed that at least one option for numberss and t corresponds to the given record.

Output

In the first line print a single number k ? the number of options for numberss and t.

In each of the following k lines print two integerssi andti ? the option for numberss and t. Print the options in the order of increasingsi, and for equalsi ? in the order of increasingti.

Sample test(s)

Input

51 2 1 2 1

Output

21 33 1

Input

41 1 1 1

Output

31 42 24 1

Input

41 2 1 2

Output

Input

82 1 2 1 1 1 1 1

Output

31 62 36 1