cf496D Tennis Game

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 of s sets, he wins the match and the match is over. Here s and t are some positive integer numbers.

To spice it up, Petya and Gena choose new numbers s and t 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 s 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 numbers s and t for the given match are also lost. The players now wonder what values of s 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 integers a_i. If a_i = 1, then the i-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 numbers s and t corresponds to the given record.

Output

In the first line print a single number k — the number of options for numbers s and t.

In each of the following k lines print two integers s_i and t_i — the option for numbers s and t. Print the options in the order of increasing s_i, and for equal s_i — in the order of increasing t_i.

Example

Input

5
1 2 1 2 1

Output

2
1 3
3 1

Input

4
1 1 1 1

Output

Input

4
1 2 1 2

Output

Input

8
2 1 2 1 1 1 1 1

Output

倍增预处理下每个数字往后2^k是哪，然后就可以logn的知道往后走n步是哪。

枚举每一个可能的“小分”，然后直接模拟下往后走。可以logn的时间知道1和2那个先到。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cstring>
  4 #include<cstdlib>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<deque>
  9 #include<set>
 10 #include<map>
 11 #include<ctime>
 12 #define LL long long
 13 #define inf 0x7ffffff
 14 #define pa pair<int,int>
 15 #define mkp(a,b) make_pair(a,b)
 16 #define pi 3.1415926535897932384626433832795028841971
 17 using namespace std;
 18 inline LL read()
 19 {
 20     LL x=0,f=1;char ch=getchar();
 21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 23     return x*f;
 24 }
 25 int n;
 26 int a[100010];
 27 int s1[100010],s2[100010];
 28 int go1[100010][20],go2[100010][20];
 29 int bin[1048577];
 30 int lst1,lst2,anst;
 31 struct aaa{int x,y;}ans[100010];
 32 bool operator <(aaa a,aaa b){return a.x<b.x;}
 33 inline int lowbit(int x){return x&(-x);}
 34 inline int calc(int s,int k,int op)
 35 {
 36     if (s==-1)return s;
 37     while (k)
 38     {
 39         if (op==1)s=go1[s][bin[lowbit(k)]];
 40         else s=go2[s][bin[lowbit(k)]];
 41         if (!s)break;
 42         k-=lowbit(k);
 43     }
 44     return s==0?-1:s;
 45 }
 46 int main()
 47 {
 48     for (int i=0;i<20;i++)bin[1<<i]=i;
 49     n=read();
 50     for (int i=1;i<=n;i++)a[i]=read();
 51     if (a[n]==2)for (int i=1;i<=n;i++)a[i]=3-a[i];
 52     for (int i=1;i<=n;i++)
 53     {
 54         s1[i]=s1[i-1]+(a[i]==1);
 55         s2[i]=s2[i-1]+(a[i]==2);
 56     }
 57     for (int i=n;i>=1;i--)
 58     {
 59         go1[i][0]=lst1;
 60         go2[i][0]=lst2;
 61         if (a[i]==1)lst1=i;
 62         else lst2=i;
 63     }
 64     for (int i=1;i<=20;i++)
 65     {
 66         if(i>n)break;
 67         for (int j=1;j<=n;j++)
 68         {
 69             if (go1[j][i-1])go1[j][i]=go1[go1[j][i-1]][i-1];
 70             if (go2[j][i-1])go2[j][i]=go2[go2[j][i-1]][i-1];
 71         }
 72     }
 73     go1[0][0]=lst1;
 74     go2[0][0]=lst2;
 75     for (int i=1;i<=20;i++)
 76     {
 77         if (i>n)break;
 78         if (go1[0][i-1]!=0)go1[0][i]=go1[go1[0][i-1]][i-1];
 79         if (go2[0][i-1]!=0)go2[0][i]=go2[go2[0][i-1]][i-1];
 80     }
 81     for (int i=1;i<=n;i++)
 82     {
 83         int cnt1=0,cnt2=0,now=0,nx1,nx2,mrk=0;
 84         while (now!=-1&&now<n)
 85         {
 86             nx1=calc(now,i,1);
 87             nx2=calc(now,i,2);
 88             if (nx1==-1&&nx2==-1){mrk=1;break;}
 89             if (nx1==-1)cnt2++,now=nx2;
 90             else if (nx2==-1)cnt1++,now=nx1;
 91             else
 92             {
 93                 if (nx1<nx2)
 94                 {
 95                     cnt1++;now=nx1;
 96                     if (now==n)break;
 97                 }else
 98                 {
 99                     cnt2++;now=nx2;
100                 }
101             }
102         }
103         if (mrk)continue;
104         if (!cnt1&&!cnt2)break;
105         if (cnt1>cnt2)ans[++anst].y=i,ans[anst].x=cnt1;
106     }
107     sort(ans+1,ans+anst+1);
108     printf("%d\n",anst);
109     for (int i=1;i<=anst;i++)printf("%d %d\n",ans[i].x,ans[i].y);
110 }

cf496D

posted on 2017-08-04 13:34 zhber 阅读(234) 评论(0) 编辑收藏举报