CCF command line options (Full Score code + problem solving ideas + skill summary) March 3, 2014

Title Description

Their thinking

• The title describes the options to output in ascending order in lowercase letters , So you can Store answers in string Array in , If the letter is an option , The output , If it is a parameterized option , Then the parameters are also output ,ans Array storage parameters
• There are two options , So first read the first line of format string into , Determine whether each letter has a reference , use Two Boolean arrays to store whether the letter has a parameter
• For the convenience of string storage , First utilize getline() Read the string contents of one line
• Again utilize stringstream Store the read one line string hierarchically in... According to the spaces vector Array in
• Then judge according to the read content , There are three situations ： No parameter option （ans[i] Store something at will ）, It is a parameterized option （ans[i] Store the next row of the array ）, Other irregularities stop parsing
• According to the ans Array to output the answer , Any options that appear on the command line , Its ans.size() It's not zero , Output option name , If it is an option with parameters , And output ans

Code implementation

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <sstream>

using namespace std;

const int N = 30;

bool op1[N], op2[N]; //op1 Store parameterless options ,op2 Parameter options are stored 
int n;
string ans[N];

int main()
{
    
    string str;
    cin >> str; // Read in all the parameters 

    for (int i = 0; i < str.size(); i ++) // Analyze whether the parameter has a parameter 
    {
    
        if (i + 1 < str.size() && str[i + 1] == ':') // If the next person doesn't cross the line , And the next digit is a colon , Then this bit is a parameterized option 
        {
    
            op2[str[i] - 'a'] = true; // indicate str[i] It is a parameterized option 
            i ++; // Skip colon 
        }
        else
        {
    
            op1[str[i] - 'a'] = true;// indicate str[i] Is an option without parameters 
        }
    }

    cin >> n;
    getchar();// Filter return 

    for (int C = 1; C <= n; C ++)
    {
    
        printf("Case %d:", C);
        getline(cin, str); // Reading the entire line requires getline, Use getline（ If the last read in used cin） You need to filter the carriage return in the previous line 
        stringstream ssin(str);
        vector <string> ops;
        while(ssin >> str) ops.push_back(str);// take str Stored hierarchically in an array by spaces 

        for (int i = 0; i < 26; i ++) ans[i].clear();// take ans Empty 

        for (int i = 1; i < ops.size(); i ++) // The first is the name , skip 
        {
    
            if (ops[i][0] != '-' || ops[i][1] < 'a' || ops[i].size() != 2) break;
            int k = ops[i][1] - 'a';
            if(op1[k])
            {
    
                ans[k] = '*';
            }
            else if(op2[k] && i + 1 < ops.size())
            {
    
                ans[k] = ops[i + 1];
                i ++;
            }
            else break;
        }

        for (int i = 0; i < 26; i ++)
        {
    
            if(ans[i].size())
            {
    
                cout << " -" << char('a' + i);
                if(op2[i])
                {
    
                    cout << ' ' << ans[i];
                }
            }
        }
        cout << endl;
    }
    return 0;
}

Skill summary

• Simulation questions are mainly Think about how to store （ Storage solves 80% of the problem ）
• A string of Selection of access mode Very important
• stringstream and vector And other common functions should be used skillfully ( The combination of the two can be Strings are stored hierarchically according to spaces ）
• string Array From the subscript 0 Start storage
• Empty string Using an array clear function
• Use words skillfully Parent subscript mapping Methods