Codeforces Round #283 (Div. 2)D（good）

D. Tennis Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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 winss 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 integersa_i. Ifa_i = 1, then thei-th serve was won by Petya, if a_i = 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 integerss_i andt_i — the option for numberss and t. Print the options in the order of increasings_i, and for equals_i — in the order of increasingt_i.

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

0

Input

82 1 2 1 1 1 1 1

Output

31 62 36 1

枚举t，然后二分每个人得分为t的位置，更新每个人赢得局数，如果符合条件，则加入答案中

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=100010;const int INF=1000000010;int sum1[maxn],sum2[maxn];vector<pair<int,int> > ansp;int N;int binary(int l,int r,int len,int *sum){    int st=l;    int ans=INF;    while(l<=r)    {        int mid=(l+r)>>1;        if(sum[mid]-sum[st-1]==len)ans=mid;        if(sum[mid]-sum[st-1]>=len)r=mid-1;        else l=mid+1;    }    return ans;}int judge(int t){    int cnt1=0,cnt2=0;    int l=1;    while(l<=N)    {        int pos1=binary(l,N,t,sum1);        int pos2=binary(l,N,t,sum2);        l=min(pos1,pos2)+1;        if(l>N+1)return 0;//没有达到t分，无法结束        if(pos1<pos2)cnt1++;        else cnt2++;        if(l-1==N)        {            if(cnt1>cnt2&&pos1>pos2)return 0;//如果第一个人赢得局数已经大于第二个人，            //而最后一局第一个人输了，说明这个s选的不合适            if(cnt1<cnt2&&pos1<pos2)return 0;        }    }    if(cnt1==cnt2)return 0;    return cnt1>cnt2?cnt1:cnt2;}int main(){    while(scanf("%d",&N)!=EOF)    {        memset(sum1,0,sizeof(sum1));        memset(sum2,0,sizeof(sum2));        for(int i=1;i<=N;i++)        {            int x;            scanf("%d",&x);            if(x==1)sum1[i]=1;            else sum2[i]=1;        }        for(int i=1;i<=N;i++)            sum1[i]+=sum1[i-1],sum2[i]+=sum2[i-1];        ansp.clear();        for(int i=1;i<=N;i++)        {            int tmp=judge(i);            if(tmp<=0)continue;            ansp.push_back(make_pair(tmp,i));        }        sort(ansp.begin(),ansp.end());        int len=ansp.size();        printf("%d\n",len);        for(int i=0;i<len;i++)            printf("%d %d\n",ansp[i].first,ansp[i].second);    }    return 0;}