Codeforces 497B Tennis Game( 枚举+ 二分)

B. 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 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 ai. If ai = 1, then the i-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 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 si and ti — the option for numbers s and t. Print the options in the order of increasingsi, and for equal si — in the order of increasing ti.

Sample test(s)

input

5
1 2 1 2 1

output

2
1 3
3 1

input

4
1 1 1 1

output

3
1 4
2 2
4 1

input

4
1 2 1 2

output

0

input

8
2 1 2 1 1 1 1 1

output

3
1 6
2 3
6 1

只要暴力枚举每盘游戏的结束分数 t， 然后判断游戏的盘数 s 。

怎么判断s呢，当然就是找s的倍数 lar 。

然后再进行二分出两个玩家拿到结束当局游戏的分数 ， 位置比较前的就是赢家 。

这题是我打那么多场DIV1最可惜的一题 ， 卡在了边缘 。没有判断第n 个位置一定是要赢家得分的 。

而且注意 ， 最后得分的人赢的游戏盘数一定比另外一个人多才符合要求 。

复杂度就是 n * log n * log n ( 枚举分数 t , 枚举 t 的倍数 ， 二分 分数的位置 ) 。

#include <bits/stdc++.h>
using namespace std;
const int N = ;
const int INF = 1e9+;
typedef pair<int,int> pii;
#define X first
#define Y second
int n , a[N], cnt1[N] , cnt2[N] ;
vector<pii>ANS;

int Find( int sc , int *A ) {

    int l =  , r = n , pos = INF ;
    if( A[n] < sc ) return pos;
    while( l <= r ){
        int mid = (l+r)>>;
        if( A[mid] < sc )
            l = mid +  ;
        else
            pos = mid , r = mid -  ;
    }
    return pos ;
}

int check( int t ) {

    int ans =  , sc1 =  , sc2 =  ;
    int c1 =  , c2 =  ;
    while(  ){
        int pos1 = Find( sc1 + t , cnt1 );
        int pos2 = Find( sc2 + t , cnt2 );
        if( pos1 == INF && pos2 == INF ) return - ;
        if( pos1 == INF ) {
            if( ( cnt2[n] - sc2 ) %t || a[n] ==  ) return - ;
            else {
                c2 += ( cnt2[n] - sc2 ) / t ;
                if( c1 >= c2 ) return - ;
                return c2 ;
            }
        }
        if( pos2 == INF ) {
            if( ( cnt1[n] - sc1 ) %t || a[n] ==  ) return - ;
            else {
                c1 += ( cnt1[n] - sc1 ) / t ;
                if( c2 >= c1 ) return - ;
                return c1 ;
            }
        }
        if( pos1 < pos2 ) c1++ , sc1 = cnt1[pos1] , sc2 = cnt2[pos1];
        else c2++ , sc1 = cnt1[pos2] , sc2 = cnt2[pos2] ;
    }
}

int main(){
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    while( cin >> n ){
        cnt1[] = cnt2[] =  ;
        for( int i =  ; i <= n ; ++i ) {
            cin >> a[i] ;
            if( a[i] ==  ) cnt1[i] = cnt1[i-]+ , cnt2[i] = cnt2[i-];
            else cnt1[i] = cnt1[i-] , cnt2[i] = cnt2[i-]+;
        }
        int tot = max( cnt1[n] , cnt2[n] ) , ans =  ;
        ANS.clear();
        for( int i =  ; i <= tot ; ++i ) {
            int res = check(i);
            if( res != - ) {
                ans++; ANS.push_back( pii( res , i ) );
            }
        }
        sort( ANS.begin() , ANS.end() ) ;
        cout << ans << endl ;
        for( int i =  ; i < ans ; ++i ) {
            cout << ANS[i].X <<' ' << ANS[i].Y << endl ;
        }
    }
}