CF1028D Order book 思维

Posted 2021-01-02 l609929321

Order book

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let‘s consider a simplified version of order book of some stock. The order book is a list of orders (offers) from people that want to buy or sell one unit of the stock, each order is described by direction (BUY or SELL) and price.

At every moment of time, every SELL offer has higher price than every BUY offer.

In this problem no two ever existed orders will have the same price.

The lowest-price SELL order and the highest-price BUY order are called the best offers, marked with black frames on the picture below.

The presented order book says that someone wants to sell the product at price $\mathrm{1212 and\; it‘s\; the\; best SELL offer\; because\; the\; other\; two\; have\; higher\; prices.\; The\; best BUY offer\; has\; price 1010.}$

There are two possible actions in this orderbook:

1. Somebody adds a new order of some direction with some price.
2. Somebody accepts the best possible SELL or BUY offer (makes a deal). It‘s impossible to accept not the best SELL or BUY offer (to make a deal at worse price). After someone accepts the offer, it is removed from the orderbook forever.

It is allowed to add new BUY order only with prices less than the best SELL offer (if you want to buy stock for higher price, then instead of adding an order you should accept the best SELL offer). Similarly, one couldn‘t add a new SELL order with price less or equal to the bestBUY offer. For example, you can‘t add a new offer "SELL $\mathrm{2020"\; if\; there\; is\; already\; an\; offer\; "BUY 2020"\; or\; "BUY 2525" —\; in\; this\; case\; you\; just\; accept\; the\; best BUY offer.}$

You have a damaged order book log (in the beginning the are no orders in book). Every action has one of the two types:

1. "ADD $\mathrm{pp"\; denotes\; adding\; a\; new\; order\; with\; price pp and\; unknown\; direction.\; The\; order\; must\; not\; contradict\; with\; orders\; still\; not\; removed\; from\; the\; order\; book.}$
2. "ACCEPT $\mathrm{pp"\; denotes\; accepting\; an\; existing\; best\; offer\; with\; price pp and\; unknown\; direction.}$

The directions of all actions are lost. Information from the log isn‘t always enough to determine these directions. Count the number of ways to correctly restore all ADD action directions so that all the described conditions are satisfied at any moment. Since the answer could be large, output it modulo ${\mathrm{109+7109+7.\; If\; it\; is\; impossible\; to\; correctly\; restore\; directions,\; then\; output 00.}}^{}$

Input

The first line contains an integer $\mathrm{nn (1\le n\le 3633041\le n\le 363304)\; —\; the\; number\; of\; actions\; in\; the\; log.}$

Each of the next $\mathrm{nn lines\; contains\; a\; string\; "ACCEPT"\; or\; "ADD"\; and\; an\; integer pp (1\le p\le 3089830661\le p\le 308983066),\; describing\; an\; action\; type\; and\; price.}$

All ADD actions have different prices. For ACCEPT action it is guaranteed that the order with the same price has already been added but has not been accepted yet.

Output

Output the number of ways to restore directions of ADD actions modulo ${\mathrm{109+7109+7.}}^{}$

Examples

input

Copy

6
ADD 1
ACCEPT 1
ADD 2
ACCEPT 2
ADD 3
ACCEPT 3

output

Copy

8

input

Copy

4
ADD 1
ADD 2
ADD 3
ACCEPT 2

output

Copy

2

input

Copy

7
ADD 1
ADD 2
ADD 3
ADD 4
ADD 5
ACCEPT 3
ACCEPT 5

output

Copy

0

Note

In the first example each of orders may be BUY or SELL.

In the second example the order with price $\mathrm{11 has\; to\; be BUY order,\; the\; order\; with\; the\; price 33 has\; to\; be SELL order.}$

 

分析：按照题目意思直接模拟

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 10;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const ll inf = 1e9;
const double pi = acos(-1.0);
int main() {
    ll n, x, le = -1e9, ri = 1e9, ans = 1, res = 1; 
    //res：区间内可买卖数量减一，初始化1方便后面运算，ans：买卖方法数，le,ri：买卖区间边界
    set<ll> s;  //买卖区间内数的集合
    set<ll>::iterator it;
    cin >> n;
    s.insert(-1e9), s.insert(1e9);
    while( n -- ) {
        string str;
        cin >> str >> x;
        if( str == "ADD" ) {
            s.insert(x);
            if( x >= le && x <= ri ) {
                res ++;
            }
        } else {
            s.insert(x);
            if( x < le || x > ri ) {
                ans = 0;
            } else if( x != le && x != ri ){
                ans = ans*2%mod;
            }
            s.erase(x); 
            res = 1; //可以买卖的区间内数为0，回到初始值1
            it = s.lower_bound(x);
            ri = *it, le = *(--it);
        }
        //debug(le), debug(ri), debug(ans);
    }
    cout << ans*res%mod << endl;
    return 0;
}

　　

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