一道纯水小模拟。

注意判定字符串合法性。递归里外都要判。
纯纯的 map 模拟，打标记，递归，只要有二级语法水平都能写。

我写了整整 15 分钟，调试整整 20 分钟，所以连二级都没有。
递归里没判合法性，直接卡我 15 分钟。

#include <bits/stdc++.h>
using namespace std;
int T;
string s;

map<string, bool> used, st;
map<string, string> dfn;

bool check(char ch) {
    if (ch >= '0' && ch <= '9') return 1;
    if (ch >= 'a' && ch <= 'z') return 1;
    if (ch >= 'A' && ch <= 'Z') return 1;
    if (ch == '_') return 1;
    return 0;
}

string solve(string str) {
    if (!used[str] || st[str]) return str;
    string p = str;
    st[str] = 1;
    str = dfn[str];
    int n = str.size();
    string s = "", ans = "";
    int id = 0;
    while (id < n) {
        while (check(str[id]) && id < n) s += str[id], id++;
        ans += solve(s), s = "";
        while (!check(str[id]) && id < n) ans += str[id], id++;
    }
    if (s.size()) ans += solve(s);
    st[p] = 0;
    return ans;
}

int main() {
    scanf("%d", &T);
    getline(cin, s);
    for (int Tid = 1; Tid <= T; Tid++) {
        getline(cin, s);
        int n = s.size();
        // cout << s << endl;
        if (s[0] == '#' && s[1] == 'd') {   //#define
            string s1 = "", s2 = ""; int pos = 0;
            for (int i = 8; i < n; i++) {
                if (s[i] == ' ') { pos = i; break; }
                s1 += s[i];
            }
            for (int i = pos + 1; i < n; i++) {
                // if (s[i] == ' ') break;
                s2 += s[i];
            }
            used[s1] = 1, dfn[s1] = s2;
            putchar('\n');
        } else if (s[0] == '#' && s[1] == 'u') {    //undef
            string s1 = "", s2 = ""; int pos = 0;
            for (int i = 7; i < n; i++) {
                if (s[i] == ' ') { pos = i; break; }
                s1 += s[i];
            }
            for (int i = pos + 1; i < n; i++) s2 += s[i];
            used[s1] = 0, dfn[s1] = "";
            putchar('\n');
        } else {
            string str = "";
            int id = 0;
            while (id < n) {
                str = "";
                while (check(s[id]) && id < n) str += s[id], id++;
                cout << solve(str);
                while (!check(s[id]) && id < n) cout << s[id], id++;
            }
            putchar('\n');
        }
    }
    return 0;
}