题目描述

源代码

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
const int N = 5000;

struct Order
{
    
    double price;
    int num;
    int sell;//交易状态：0-买入 1-卖出 -1 取消
} order[N + 1];

vector<double> pricevec;//Save all occurrences of prices

int main()
{
    
    char s[10];
    int num = 0, t; //numThe total number of recorded data

    while (cin >> s) //buy 9.25 100
    {
    
        if (s[0] == 'c')//读入的是cancel
        {
    
            cin >> t;
            order[t - 1].sell = -1;//取消第iline of transactions
            order[num].sell = -1;//cancelInstructions must also occupy one
        }
        else
        {
    
            cin >> order[num].price >> order[num].num;
            if (s[0] == 's')//卖出
                order[num].sell = 1;
            else
                order[num].sell = 0;
            pricevec.push_back(order[num].price);
        }
        num++;
    }

    double ans = 0;//开盘价
    long long maxn = 0;//Save the maximum transaction volume
    for (int i = 0; i < pricevec.size(); i++)
    {
    
        long long num_s = 0;//The number of shares available for sale
        long long num_b = 0;//The number of shares available for purchase

        for (int j = 0; j < num; j++)
        {
    
            if (order[j].sell == 1 && order[j].price <= pricevec[i])
                num_s += (long long)order[j].num;
        }

        for (int j = 0; j < num; j++)
        {
    
            if (order[j].sell == 0 && order[j].price >= pricevec[i])
                num_b += (long long)order[j].num;
        }

        num_s = min(num_s, num_b);

        if (num_s > maxn)
        {
    
            maxn = num_s;
            ans = pricevec[i];
        }
        else
        {
    
            if (num_s == maxn)
                ans = max(ans, pricevec[i]);
        }
    }

    printf("%.2f %lld\n", ans, maxn);

    return 0;
}