SMU 2024 spring 天梯赛2

SMU 2024 spring 天梯赛2

7-1 计算指数 - SMU 2024 spring 天梯赛2 (pintia.cn)

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n,ans = 1;
    cin >> n;
    for(int i = 1;i <= n;i ++, ans *= 2);
    cout << "2^"<< n<<" = " << ans;

    return 0;
}

7-2 计算摄氏温度 - SMU 2024 spring 天梯赛2 (pintia.cn)

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cout << "fahr = 100, celsius = "<<5 * (100-32)/9 ;

    return 0;
}

7-3 念数字 - SMU 2024 spring 天梯赛2 (pintia.cn)

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    vector<string> a = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
    string s;
    cin >> s;
    if(s[0] == '-') cout << "fu ",s.erase(s.begin());

    for(int i = 0;i < s.size();i++)
        cout << a[s[i] - '0'] << " \n"[i == s.size() - 1];

    return 0;
}

7-4 求阶乘累加和 - SMU 2024 spring 天梯赛2 (pintia.cn)

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    i64 n, ans = 0,a = 1;
    cin >> n;
    for(int i = 1;i <= n;i ++){
        a *= i;
        ans += a;
    }

    cout << ans << '\n';

    return 0;
}

7-5 6翻了 - SMU 2024 spring 天梯赛2 (pintia.cn)

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string s;
    getline(cin, s);
    for (int i = 0; i < s.size();) {
        if (s[i] == '6') {
            int j = i;
            while (j < s.size() && s[j]  == '6') j ++;
            if (j - i > 9) cout << "27", i = j;
            else if (j - i > 3) cout << "9", i = j;
            else cout << s[i] , i ++;
        } else
            cout << s[i], i ++;
    }

    return 0;
}

7-6 福到了 - SMU 2024 spring 天梯赛2 (pintia.cn)

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    char c;
    int n;
    cin >> c >> n;
    vector<string> a(n), b(n);
    string s;
    getline(cin,s);
    for (int i = 0;i < n;i ++) {
        getline(cin,s);
        b[i] = a[i] = s;
        reverse(b[i].begin(), b[i].end());
    }

    reverse(b.begin(), b.end());
    if (a == b)  cout << "bu yong dao le\n";

    for (auto s : b) {
        for (auto j : s)
            if (j == '@') cout << c;
            else cout << j;
        cout << '\n';
    }

    return 0;
}

7-7 估值一亿的AI核心代码 - SMU 2024 spring 天梯赛2(补题) (pintia.cn)

很恶心的模拟题,做一遍就不想做第二遍;

#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;
    string s;
    getline(cin, s);

    auto biaoDian = [](char c) -> bool{
        if (c >= 'A' && c <= 'Z') return false;
        if (c >= 'a' && c <= 'z') return false;
        if (c >= '0' && c <= '9') return false;
        return true;
    };

    for (int i = 0; i < n; i ++) {
        getline(cin, s);
        cout << s << '\n';
        vector<string> a;
        int j = 0, k = 0;
        for (int l = 0; l < s.size(); l ++)
            if (s[l] == '?') s[l] = '!';
        while (k < s.size()) {
            if (s[k] == ' ') {
                while (k + 1 < s.size() && s[k + 1] == ' ') k ++;
                if (a.empty()) {
                    k ++;
                    j = k;
                    continue;
                };
                a.push_back(" ");
            } else if (!biaoDian(s[k])) {
                while (k + 1 < s.size() && s[k + 1] != ' ' && !biaoDian(s[k + 1])) k ++;
                a.push_back(s.substr(j, k - j + 1));
            } else {
                string as = "";
                as += s[k];
                a.push_back(as);
            }
            k ++;
            j = k;
        }

        if (a.size() && a.back() == " ") a.pop_back();
        for (j = a.size() - 1; j > 0; j --) {
            if (biaoDian(a[j][0])) {
                if (a[j - 1] == " ") a.erase(a.begin() + j - 1);
            }
        }

        for (j = 0; j < a.size(); j ++) {
            for (k = 0; k < a[j].size(); k ++) {
                if (a[j][k] >= 'A' && a[j][k] < 'Z' && a[j][k] != 'I')
                    a[j][k] += 32;
            }
        }

        for (j = 0; j < a.size(); j ++) {
            if (a[j] == "I" || a[j] == "me") {
                a[j] = "you#";
                continue;
            }
        }

        for (j = 0; j < a.size(); j ++) {
            if (j + 2 >= a.size()) continue;
            string p = a[j] + a[j + 1] + a[j + 2];
            if (p == "can you" || p == "could you") {
                a[j + 2] = a[j];
                a[j] = "I";
                j += 2;
            }
        }

        string ans = "";
        for (auto p : a)
            ans += p;
        cout << "AI: ";
        for (j = 0; j < ans.size(); j ++)
            if (ans[j] == '#') continue;
            else cout << ans[j];
        cout << '\n';

    }

    return 0;
}

7-8 前世档案 - SMU 2024 spring 天梯赛2(补题) (pintia.cn)

是对照0,否对照1;

按照上述去判断对应问题即结论1为000,结论2为001 \(\dots\) 结论8为111;

很明显可以用递归去判断,结论数即对应的二进制转十进制后加1即可;

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m;
    cin >> n >> m;

    while (m --) {
        string s;
        cin >> s;
        auto dfs = [&](auto dfs, int now) -> i64{
            if (now >= n) return 1;
            if (s[now] == 'y')
                return dfs(dfs, now + 1);
            else
                return (1ll << (n - now - 1)) + dfs(dfs, now + 1);
        };
        cout << dfs(dfs, 0) << '\n';
    }

    return 0;
}

7-9 抢红包 - SMU 2024 spring 天梯赛2(补题) (pintia.cn)

排序,注意最后输出保留两位小数;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

struct people{
	int id,num = 0;
	double val = 0;
};

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

 	int n;
 	cin >> n;
 	vector<people> Ke(n + 1);
 	for(int i = 1;i <= n;i ++){
 		int k,num = 0;
 		cin >> k;
 		Ke[i].id = i;
 		for(int j = 0;j < k;j ++){
 			int ni,pi;
 			cin >> ni >> pi;
 			num += pi;
 			Ke[ni].num ++;
 			Ke[ni].val += pi;
 		}
 		Ke[i].val -= num;
 	}

 	Ke.erase(Ke.begin());
 	sort(Ke.begin(),Ke.end(),[&](people a,people b){
 		if(a.val == b.val && a.num == b.num) return a.id < b.id;
 		if(a.val == b.val) return a.num > b.num;
 		return a.val > b.val;
 	});

 	for(auto [id,num,val] : Ke){
 		printf("%d %.2lf\n",id, val / 100);
 	}

	return 0;
}

7-10 红色警报 - SMU 2024 spring 天梯赛2(补题) (pintia.cn)

并查集;

先统计原来的联通快有多少;

每次去掉一个城市后用并查集统计联通块数与上次是否发生变化;

小于上次是有可能有些单独的点被移除了,但是并不影响其余联通块;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

struct UFS {
	vector<int> fa;
	int n;
	UFS(int n) : n(n) {
		fa.resize(n + 1);
		for (int i = 0; i <= n; i ++)
			fa[i] = i;
	}

	int find(int x) {
		return fa[x] == x ? x : find(fa[x]);
	}

	void unin(int x, int y) {
		x = find(x), y = find(y);
		if (x != y) fa[x] = y;
	}
};

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n, m;
	cin >> n >> m;

	typedef pair<i64, i64> PII;
	int num = 0;
	vector<PII> edge(m);
	UFS ufs(n);
	for (auto &[u, v] : edge) {
		cin >> u >> v;
		ufs.unin(u, v);
	}

	for (int i = 0; i < n; i ++)
		if (ufs.find(i) == i) num ++;

	int k;
	cin >> k;
	vector<bool> vis(n);
	while (k --) {
		int x, now = 0;
		cin >> x;
		UFS ufs1(n);
		vis[x] = 1;
		for (int i = 0; i < m; i ++) {
			auto [u, v] = edge[i];
			if (vis[u] || vis[v]) continue;
			ufs1.unin(u, v);
		}
		for (int i = 0; i < n; i ++)
			if (ufs1.find(i) == i && !vis[i]) now ++;
		if (now <= num) {
			cout << "City " << x << " is lost.\n";
		} else {
			cout << "Red Alert: City " << x << " is lost!\n";
		}
		num = now;
		bool f = false;
		for (int i = 0; i < n; i ++)
			if (!vis[i]) f = true;
		if (!f) cout << "Game Over.\n";
	}

	return 0;
}

7-11 排座位 - SMU 2024 spring 天梯赛2(补题) (pintia.cn)

并查集;

用并查集统计朋友,敌人用二维关系表示即可,因为没有说明敌人的敌人也是朋友,所以不适合种类并查集;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

struct UFS {
	vector<int> fa;
	int n;
	UFS(int n) : n(n) {
		fa.resize(n + 1);
		for (int i = 0; i <= n; i ++)
			fa[i] = i;
	}
	int find(int x) {
		return fa[x] == x ? x : fa[x] = find(fa[x]);
	}
	void unin(int x, int y) {
		x = find(x), y = find(y);
		if (x != y) fa[x] = y;
	}
};

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n, m, k;
	cin >> n >> m >> k;
	UFS frend(n);
	vector Diren(n + 1,vector<int>(n + 1));
	for (int i = 0; i < m; i ++) {
		int u, v, c;
		cin >> u >> v >> c;
		if (c == 1) {
			frend.unin(u, v);
		} else Diren[u][v] = 1,Diren[v][u] = 1;
	}

	while(k --){
		int u,v;
		cin >> u >> v;
		int a = frend.find(u),b = frend.find(v);
		if(a == b && !Diren[u][v]){
			cout << "No problem\n";
		}else if(a != b && !Diren[u][v]){
			cout << "OK\n";
		}else if(a == b && Diren[u][v]){
			cout << "OK but...\n";
		}else{
			cout << "No way\n";
		}
	}

	return 0;
}

7-12 这是二叉搜索树吗？ - SMU 2024 spring 天梯赛2(补题) (pintia.cn)

建一颗二叉搜索树,跑一遍dfs记录前后序遍历;

如果是镜像就在原来的dfs基础上交换一下左右子树的顺序即可;

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

struct Node {
	int data;
	Node* left;
	Node* right;

	Node* newNode(int v) {
		Node* node = new Node;
		node->data = v;
		node->right = node->left = NULL;
		return node;
	}

	void insert(Node* &root, int x) {
		if (root == NULL) {
			root = newNode(x);
			return ;
		}
		if (root->data > x)
			insert(root->left, x);
		else
			insert(root->right, x);
	}

	Node* Create(vector<int> tree) {
		Node* root = NULL;
		for (auto i : tree) {
			insert(root, i);
		}
		return root;
	}
};

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;
	vector<int> tree(n);
	for (auto &i : tree) cin >> i;

	Node Tr;

	vector<int> pre, suf, Jpre;
	auto dfs = [&](auto dfs, Node * root) {
		if (root == NULL) return ;
		pre.push_back(root->data);
		dfs(dfs, root->left);
		dfs(dfs, root->right);
		suf.push_back(root->data);
	};

	dfs(dfs, Tr.Create(tree));

	if (pre == tree) {
		cout << "YES\n";
		for (int i = 0; i < suf.size(); i ++)
			cout << suf[i] << " \n"[i == suf.size() - 1];
		return 0;
	}

	vector<int>().swap(suf);
	auto dfs1 = [&](auto dfs, Node * root) {
		if (root == NULL) return ;
		Jpre.push_back(root->data);
		dfs(dfs, root->right);
		dfs(dfs, root->left);
		suf.push_back(root->data);
	};

	dfs1(dfs1, Tr.Create(tree));
	if (Jpre == tree) {
		cout << "YES\n";
		for (int i = 0; i < suf.size(); i ++)
			cout << suf[i] << " \n"[i == suf.size() - 1];
		return 0;
	}

	cout << "NO\n";

	return 0;
}