2022网易雷火游戏研发笔试ak（4月23日）

首先，可以用自己的IDE！！！模拟体验满分！！！

下午刚打完天梯赛，被模拟折磨了一下午，晚上又被模拟折磨惹。
40分钟（好像是）写完ABC，看了看D发现是阅读理解，本来准备交卷开摆，后来想了想，多交几发枚举一下题意试试，发现和天梯赛的模拟难度差不多。。。

A题：签到，怎么写都能过吧

B题：签到，直接dp即可

C题：他让你怎么写你就怎么写

#include <bits/stdc++.h>
using namespace std;

#define ll long long

struct func {
	string name;
	string argdf;
	vector<string>s;
	vector<int>i;
}fun[1000];

map<char, int>htd;

string ITS(int x) {
	string res="";
	if (!x)return "0";
	bool flag = 1;
	if (x < 0)
		flag = 0, x = -x;
	while (x) {
		res = char((x % 10) + '0') + res;
		x /= 10;
	}
	if (!flag)
		res = '-' + res;
	return res;
}

int main() 
{
	for (int i = 0; i <= 9; i++)
		htd[char(i + '0')] = i;
	htd['A'] = 10;
	htd['B'] = 11;
	htd['C'] = 12;
	htd['D'] = 13;
	htd['E'] = 14;
	htd['F'] = 15;

	int n;
	cin >> n;
	for (int i = 1; i <= n;i++) {
		int x;
		cin >> x;
		string name, argd;
		cin >> name >> argd;
		fun[x].name = name;
		fun[x].argdf = argd;
	}

	string rpcs="";
	string tmp;
	while (cin >> tmp) {
		rpcs += tmp;
	}

	int now = 0;
	for (; now < rpcs.length();) {
		int rpcid = 0;
		rpcid = htd[rpcs[now]] * 16 + htd[rpcs[now + 1]];
		assert(rpcid < 256);
		now += 2;

		string name = fun[rpcid].name;
		int tot = 0;
		for (auto i:fun[rpcid].argdf) {
			if (i == 's')
				tot++;
		}


		queue<int>len;
		for (int i = 1; i <= tot; i++) {
			len.push(htd[rpcs[now]] * 16 + htd[rpcs[now + 1]]);
			now += 2;
		}

		for (auto o : fun[rpcid].argdf) {
			if (o == 's') {
				string tmp = "";
				int i = len.front();
				len.pop();
				for (int j = 1; j <= i * 2; j++) {
					tmp += rpcs[now++];
				}
				fun[rpcid].s.push_back(tmp);
			}
			else {
				int sum = 0;
				for (int j = 1; j <= 8; j++) {
					sum = sum * 16 + htd[rpcs[now++]];
				}
				fun[rpcid].i.push_back(sum);
			}
		}

		string ans = name + "(";
		int pt1 = 0, pt2 = 0;
		for (auto i : fun[rpcid].argdf) {
			if (i == 's') {
				ans += "\"";
				ans += fun[rpcid].s[pt1++];
				ans += "\"";
			}
			else {
				ans += ITS(fun[rpcid].i[pt2++]);
			}
			ans += ",";
		}
		fun[rpcid].s.clear();
		fun[rpcid].i.clear();
		ans.pop_back();
		ans += ")";
		cout << ans << endl;
	}


	return 0;
}

D题：同上，需要一点点技巧。由于输入已经分好词了，所以对于每条question，遍历每个规则，用栈匹配处理括号，同时对于规则里每个词都去这条question里找存不存在，就可以把规则中的单词变成0和1，然后规则就变成只含\(+\)、\(*\)和括号的布尔表达式求值

剩下的就是他怎么说你怎么写了。

#include <bits/stdc++.h>
using namespace std;

#define ll long long

map<string, int>id;

set<int>q[50];

vector<string>ans;

string getNextWord(string& s, int& i) {
	if (i >= s.length())return "";
	while (s[i] == ' ')
		i++;
	string res = "";
	while (i < s.length() && s[i] != ' ')
		res = res + s[i++];
	return res;
}

int main()
{
	int n, m;
	cin >> n >> m;
	string s;
	getline(cin, s);
	int cnt = 0;
	for (int i = 1; i <= n; i++) {
		getline(cin, s);
		string tmp = "";
		for (int j = 0; j < s.length(); j++) {
			if (s[j] == ' ') {
				if (id.find(tmp) == id.end())
					id[tmp] = ++cnt;
				q[i].insert(id[tmp]);
				tmp.clear();
				continue;
			}
			tmp += s[j];
		}
		if (id.find(tmp) == id.end())
			id[tmp] = ++cnt;
		q[i].insert(id[tmp]);
	}

	for (int i = 1; i <= m; i++) {
		getline(cin, s);
		ans.push_back(s);
	}


	for (int qi = 1; qi <= n; qi++) {
		auto que = q[qi];
		bool ok = 0;
		int mx = 0;
		string res;
		for (auto as : ans) {
			int i = 0;
			stack<int>stk;
			int flag = 0;
			string now = "";
			int tot = 0;//关键词数量
			while (1) {
				now = getNextWord(as, i);
				if (now == "")break;
				if (now == "and") {
					flag = 1;
				}
				else if (now == "or") {
					flag = 0;
				}
				else if (now == ")") {
					while (1) {
						int x = stk.top();
						stk.pop();
						if (stk.top() == 114 || stk.top() == 514) {
							if (stk.top() == 114) {
								flag = 0;
								stk.pop();
								stk.push(x);
							}
							else {
								flag = 1;
								stk.pop();
								stk.top() &= x;
							}
							break;
						}
						stk.top() |= x;
					}
				}
				else if (now == "(") {
					if (flag == 0)
						stk.push(114);
					else 
						stk.push(514);
					flag = 0;
					continue;
				}
				else {
					tot++;
					int x = 0;
					if (id.find(now) == id.end()) {
						x = 0;
					}
					else {
						if (que.find(id[now]) != que.end()) {
							x = 1;
						}
						else {
							x = 0;
						}
					}
					if (flag == 0) {
						stk.push(x);
					}
					else {
						stk.top() &= x;
					}
				}
			}
			while (stk.size() > 1) {
				int x = stk.top();
				stk.pop();
				stk.top() |= x;
			}
			if (stk.top() == 1) {
				if (tot > mx) {
					mx = tot;
					res = as;
				}
				ok = 1;
			}
		}
		if (!ok)
			cout << "miss" << endl;
		else
			cout << res << endl;
	}


	return 0;
}