CCF call auction (full mark code + problem solving ideas + skill summary) 201412 - 3

Skill summary

• When you need a pair of data to get the most value , It is not necessary to use pair For storage （ More complicated ）, Maintain the current maximum value with two variables
• Pay attention to whether the data range will detonation int
• All data must be defined Don't forget to initialize

Title Description

Their thinking

• The data range is not large, only 5000, So you can choose to use violence n*2 solve
• As the title indicates the same opening volume , Choose a higher opening price , therefore The enumerated price should be the price given in the title , Because if you take the middle of the two prices in the title , Then the highest opening price of the same opening trading volume must be the price already given
• The meaning of the title is concise and comprehensive , For one Given price p, Calculate purchase price >= p Number of shares , Selling price <= p Number of shares , both Take the smaller value , Calculate the maximum of all transaction volumes , And the corresponding maximum price p
  
  

Code implementation

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 5010;

struct share
{
    
    string character;
    double price;
    int number;
    bool flag;
}shares[N];

int main()
{
    
    int cnt = 0;
    string a;
    while(cin >> a)
    {
    
        double p = 0.0;
        int n = 0;
        int c = 0;
        if (a == "buy")
        {
    
            cin >> p >> n;
            shares[ ++ cnt] = {
    "buy", p, n, true};
        }
        else if(a == "sell")
        {
    
            cin >> p >> n;
            shares[ ++ cnt] = {
    "sell", p, n, true};
        }
        else if (a == "cancel")
        {
    
            cin >> c;
            shares[c].flag = false;
            shares[ ++ cnt].flag = false;
        }
    }

    long long s = 0; // Be sure to initialize 
    double p = 0;
    for (int i = 1; i <= cnt; i ++)
    {
    
        if(!shares[i].flag) continue;
        long long sum1 = 0;
        long long sum2 = 0;
        for (int j = 1; j <= cnt; j ++)
        {
    
            if(!shares[j].flag) continue;
            if(shares[j].character == "buy")
            {
    
                if(shares[j].price >= shares[i].price) sum1 += shares[j].number;
            }
            else if(shares[j].character == "sell")
            {
    
                if(shares[j].price <= shares[i].price) sum2 += shares[j].number;
            }
        }
        long long res = min(sum1, sum2);
        if(res > s || (res == s && shares[i].price > p))
        {
    
            s = res;
            p = shares[i].price;
        }
    }

    printf("%.2f", p);
    cout << " " << s;
    return 0;
}