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📂SMU2025训练记录
🔖模拟树
2025-01-26 10:57阅读: 8评论: 0推荐: 0

2025寒假天梯赛训练1

2025寒假天梯赛训练1

7-1 心理阴影面积

思路

用一半的面积减去一个梯形和一个三角形面积即可。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int a, b;
	cin >> a >> b;
 
	cout << (5000 - (100 - a) * 50 - b * 50) << "\n";
 
	return 0;
}

7-2 人与神

思路

直接输出即可。

代码

To iterate is human, to recurse divine.

7-3 通讯录的录入与显示

思路

按题意输出即可。

注意记录编号可能为负数。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
 
	vector<array<string,5>> s(n);
	for(auto &x : s){
		for(int i = 0;i < 5;i ++){
			cin >> x[i];
		}
	}	
 
	int k;
	cin >> k;
 
	while(k--){
		int x;
		cin >> x;
        
		if(x >= n || x < 0){
			cout << "Not Found\n";
		}else{
			cout << s[x][0] << " " << s[x][3] << " " << s[x][4] << " " << s[x][2] << " " << s[x][1] << "\n";
		}
	}
 
	return 0;
}

7-4 算术入门之加减乘除

思路

按题意要求输出即可。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int a,b;
	cin >> a >> b;
 
	cout << a << " + " << b << " = " << a + b << "\n"	;
	cout << a << " - " << b << " = " << a - b << "\n"	;
	cout << a << " * " << b << " = " << a * b << "\n"	;
	if(a % b == 0){
		cout << a << " / " << b << " = " << a / b << "\n"	;
	}else{
		cout << fixed << setprecision(2) << a << " / " << b << " = " << a * 1.0 / b << "\n"	;
	}
 
	return 0;
}

7-5 出生年

思路

按题意模拟。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int a,b;
	cin >> a >> b;
 
	int cs = a;
	while(1){
		set<int> has;
		if(a < 1000){
			has.insert(0);
		}
		int t = a;
		while(t){
			has.insert(t%10);
			t/=10;
		}
		if(has.size() == b){
			printf("%d %04d\n",a-cs,a);
			break;
		}
		a ++;
	}
 
	return 0;
}

7-6 九宫格输入法

思路

按题意模拟。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	vector<string> s{
		"0 ",
		"1,.?!",
		"2ABC",
		"3DEF",
		"4GHI",
		"5JKL",
		"6MNO",
		"7PQRS",
		"8TUV",
		"9WXYZ"
	};
 
	string x;
	while (cin >> x) {
		int a = x[0] - '0';
		int b = x.size();
 
		b--;
		b %= (int)s[a].size();
 
		cout << s[a][b];
	}
 
	return 0;
}

7-7 螺旋方阵

思路

按题意模拟。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
 
	vector a(n + 1, vector(n + 1, 0));
 
	int has = 0, x = 1, y = 0;
	while (has < n * n) {
		while(y + 1 <= n && !a[x][y + 1]){
			y ++;
			a[x][y] = ++has;
		}
		while(x + 1 <= n && !a[x + 1][y]){
			x ++;
			a[x][y] = ++has;
		}
		while(y - 1 >= 1 && !a[x][y - 1]){
			y --;
			a[x][y] = ++has;
		}
		while(x - 1 >= 1 && !a[x-1][y]){
			x --;
			a[x][y] = ++has;
		}
	}
 
	for(int i = 1;i <= n;i ++){
		for(int j = 1;j <= n;j ++){
			cout << setw(3) << a[i][j];
		}
		cout << "\n";
	}
 
	return 0;
}

7-8 抓老鼠啊~亏了还是赚了？

思路

有点需要注意细节的模拟。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	string s;
	cin >> s;
 
	int happy = 0, nohappy = 0, sad = 0;
	int nai = 0, now = 0;
	string ans = "";
 
	auto clear = [&]() {
		happy--;
		happy = max(happy, 0);
		nohappy--;
		nohappy = max(nohappy, 0);
		sad--;
		sad = max(sad, 0);
	};
 
	for (int i = 0; i < s.size() - 1; i ++) {
		if (s[i] == 'X') {
			if (happy || (!nohappy && !sad)) {
				ans += "U";
				nohappy = 2;
			} else {
				ans += "-";
			}
		} else if (s[i] == 'T') {
			if (happy || (!nohappy && !sad)) {
				ans += "D";
				now += 7;
				sad = 3;
			} else {
				ans += "-";
			}
		} else {
			if (happy || (!nohappy && !sad)) {
				ans += "!";
				now -= 3;
				happy = 3;
			} else {
				ans += "-";
			}
		}
		clear();
	}
 
	cout << ans << "\n" << now << "\n" ;
 
	return 0;
}

7-9 Windows消息队列

思路

用优先队列维护优先级即可。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
 
	priority_queue<pair<int, string>,vector<pair<int, string>>,greater<pair<int, string>>> Q;
	while (n--) {
		string x;
		cin >> x;
		if (x == "GET") {
			if (Q.empty()) {
				cout << "EMPTY QUEUE!\n";
			} else {
				auto [_, s] = Q.top();
				Q.pop();
				cout << s << "\n";
			}
		} else {
			int m;
			cin >> x >> m;
			Q.emplace(m, x);
		}
	}
 
	return 0;
}

7-10 名人堂与代金券

思路

考察排序。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n,g,k;
	cin >> n >> g >> k;
 
	vector<pair<string,int>> a(n);
 
	int need = 0;
	for(auto &[x,y] : a){
		cin >> x >> y;
		if(y >= g){
			need += 50;
		}else if(y >= 60){
			need += 20;
		}
	}	
 
	cout << need << "\n";
 
	sort(a.begin(),a.end(),[](auto x,auto y){
		if(x.second != y.second) return x.second > y.second;
		return x.first < y.first;
	});
 
	int rank = 1;
	for(int i = 0;i < n;i ++){
		if(rank > k) break;
		cout << rank << " " << a[i].first << " " << a[i].second << "\n";
		while(i + 1 < n && a[i + 1].second == a[i].second){
			i ++;
			cout << rank << " " << a[i].first << " " << a[i].second << "\n";
		}
		rank = i + 2;
	}
 
	return 0;
}

7-11 用扑克牌计算24点

思路

答辩题。

实际要注意的细节比较多，括号的顺序可以直接列出来然后挨个判断，其余的就是暴力搜索了。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	string a[4];
	for (int i = 0; i < 4; i ++) {
		cin >> a[i];
	}
 
	auto check = [&](string & s)->bool{
		int i = 0;
 
		auto calc = [](double x, double y, char op)->double{
			if (op == '+') return x + y;
			if (op == '-') return x - y;
			if (op == '*') return x * y;
			return x * 1.0 / y;
		};
 
		auto Calc = [&](auto && self)->double{
			char op = '+';
			stack<double> st;
			for (; i < s.size(); i++) {
				if (s[i] == '(') {
					i ++;
					st.push(self(self));
				} else if (s[i] != ')') {
					if (isdigit(s[i])) {
						int x = s[i] - '0';
						if (isdigit(s[i + 1])) {
							x = x * 10 + s[i + 1] - '0';
							i ++;
						}
						st.push(x);
					} else {
						op = s[i];
					}
				} else {
					double y = st.top();
					st.pop();
					double x = st.top();
					st.pop();
					return calc(x, y, op);
				}
			}
			if (st.size() > 1) {
				double y = st.top();
				st.pop();
				double x = st.top();
				st.pop();
				st.push(calc(x, y, op));
			}
			return st.top();
		};
 
		bool ok = Calc(Calc) == 24;
		return ok;
	};
 
	string p = "+-*/";
 
	string s = "((2-(3+12))+12)";
 
	sort(a, a + 4);
	do {
		for (auto x : p) {
			for (auto y : p) {
				for (auto z : p) {
					s = "((" + a[0] + x + a[1] + ")" + y + a[2] + ")" + z + a[3];
					if (check(s)) {
						cout << s << "\n";
						return 0;
					}
					s = "(" + a[0] + x + a[1] + ")" + y + "(" + a[2] + z + a[3] + ")";
					if (check(s)) {
						cout << s << "\n";
						return 0;
					}
					s = a[0] + x + "(" + a[1] + y + "(" + a[2] + z + a[3] + "))" ;
					if (check(s)) {
						cout << s << "\n";
						return 0;
					}
					s = a[0] + x + "((" + a[1] + y + a[2] + ")" + z + a[3] + ")";
					if (check(s)) {
						cout << s << "\n";
						return 0;
					}
					s = "(" + a[0] + x + "(" + a[1] + y + a[2] + "))" + z + a[3];
					if (check(s)) {
						cout << s << "\n";
						return 0;
					}
				}
			}
		}
	} while (next_permutation(a, a + 4));
 
	cout << "-1\n";
 
	return 0;
}

7-12 玩转二叉树

思路

先用中序前序按层还原二叉树，然后对于每一层反向输出即可。

代码

#include <bits/stdc++.h>
 
using namespace std;
 
using i64 = long long;
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
 
	int n;
	cin >> n;
 
	vector<int> pre(n), mid(n);
	for (int i = 0; i < n; i ++) {
		cin >> mid[i];
	}
	for (int i = 0; i < n; i ++) {
		cin >> pre[i];
	}
 
	map<int, vector<int>> tr;
 
	int len = 0;
	auto dfs = [&](auto && self, int rt, int st, int ed, int len)->void{
		if (st > ed) {
			return;
		}
 
		int root = st;
		while (root < ed && pre[rt] != mid[root]) {
			root ++;
		}
 
		self(self, rt + 1, st, root - 1, len + 1);
		self(self, rt + root - st + 1, root + 1, ed, len + 1);
		tr[len].emplace_back(pre[rt]);
	};
 
	dfs(dfs, 0, 0, n - 1, 0);
 
	vector<int> ans;
	for (int i = 0; i < n; i ++) {
		if (tr[i].size()) {
			for (int j = tr[i].size() - 1; j >= 0; j --) {
				ans.push_back(tr[i][j]);
			}
		}
	}
 
	for (int i = 0; i < n; i ++) {
		cout << ans[i] << " \n"[i == n - 1];
	}
 
	return 0;
}

本文作者：Ke_scholar

本文链接：https://www.cnblogs.com/Kescholar/p/18691586

版权声明：本作品采用知识共享署名-非商业性使用-禁止演绎 2.5 中国大陆许可协议进行许可。

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