Mars Numbers

题目链接：1100 Mars Numbers (20 point(s))

题干大意

给定规则的进制转换

思路

起初我是打算输入一个处理一个的，但下手之前看到了《算法笔记》中的思路，直接用map做（因为数据量也不大）完全可以预处理，然后后面直接查询

AC代码

#include <iostream>
#include <map>
#include <string>
#include <vector>
using namespace std;

const vector<string> unit = {"tret", "jan", "feb", "mar", "apr", "may", "jun",
                             "jly",  "aug", "sep", "oct", "nov", "dec"}; // NOLINT
const vector<string> ten  = {"tret", "tam", "hel", "maa", "huh", "tou", "kes",
                             "hei",  "elo", "syy", "lok", "mer", "jou"}; // NOLINT

int main()
{
    map<string, int> str2num;
    map<int, string> num2str;
    for (int i = 0; i < 13; ++i) {
        num2str[i]       = unit[i]; // 个位为[0, 12]，十位为0
        str2num[unit[i]] = i;
        num2str[i * 13]  = ten[i]; // 十位为[0, 12], 个位为9
        str2num[ten[i]]  = i * 13;
    }
    for (int i = 1; i < 13; ++i) {
        for (int j = 1; j < 13; ++j) {
            string str          = ten[i] + " " + unit[j];
            num2str[i * 13 + j] = str;
            str2num[str]        = i * 13 + j;
        }
    }
    int n;
    cin >> n;
    cin.get();
    while (n--) {
        string str;
        getline(cin, str);
        if (str[0] >= '0' && str[0] <= '9') { // 是数字
            int num = stoi(str);
            cout << num2str[num] << endl;
        } else {
            cout << str2num[str] << endl;
        }
    }
    return 0;
}