SMU 2024 spring 天梯赛4

SMU 2024 spring 天梯赛4

7-1 心理阴影面积 - SMU 2024 spring 天梯赛4 (pintia.cn)

由 \(d = \frac{Ax+By+c}{\sqrt {A^2 + B^2}}\) 再乘以 \(100\sqrt 2 \times \frac{1}{2}\)可得\(|x-y| \times 50\);

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

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

	int x,y;
	cin >> x >> y;
    cout << abs(y-x) * 50 << '\n';

	return 0;
}

7-2 人与神 - SMU 2024 spring 天梯赛4 (pintia.cn)

To iterate is human, to recurse divine.

7-3 通讯录的录入与显示 - SMU 2024 spring 天梯赛4 (pintia.cn)

要查询的记录编号可能为负数;

#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;
	vector<array<string,5>> mp(n + 1);
	for(int i = 0;i < n;i ++){
		for(int j = 0;j < 5;j ++)
			cin >> mp[i][j];
	}

	int k;
	cin >> k;
	while(k--){
		int num;
		cin >> num;
		if(num < n && num >= 0){
			cout << mp[num][0] << ' ' << mp[num][3] << ' ' << mp[num][4] << ' ' << mp[num][2] << ' ' << mp[num][1] << '\n';
		}else{
			cout << "Not Found\n";
		}
	}
	return 0;
}

7-4 算术入门之加减乘除 - SMU 2024 spring 天梯赛4 (pintia.cn)

#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;
    printf("%d + %d = %d\n",a,b,a+b);
    printf("%d - %d = %d\n",a,b,a-b);
    printf("%d * %d = %d\n",a,b,a*b);
    if(a % b == 0)
        printf("%d / %d = %d\n",a,b,a/b);
    else
        printf("%d / %d = %.2lf\n",a,b,a * 1.0 / b);

    return 0;
}

7-5 出生年 - SMU 2024 spring 天梯赛4 (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,x,y;
	cin >> y >> n;

	for(int i = y;;i ++){
		set<int> s;
		if(i<1000) s.insert(0);
		int j = i;
		while(j){
			s.insert(j % 10);
			j /= 10;
		}
		if(s.size() == n){
			printf("%d %04d\n",i - y, i);
			break;
		}
	}

	return 0;
}

7-6 九宫格输入法 - SMU 2024 spring 天梯赛4 (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> ok{
        {"0 "},{"1,.?!"},{"2ABC"},{"3DEF"},{"4GHI"},{"5JKL"},{"6MNO"},{"7PQRS"},{"8TUV"},{"9WXYZ"}
    };

    string s;
    getline(cin,s);
    for(int i = 0;i < s.size();i ++){
        if(s[i] != ' '){
            int j = i;
            while(j + 1 < s.size() && s[j + 1] != ' ') j ++;
            int num = j - i;
            string p = ok[s[i] - '0'];
            cout << p[num % (int)p.size()];
            i = j + 1;
        }
    }

    return 0;
}

7-7 螺旋方阵 - SMU 2024 spring 天梯赛4 (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;
    cin >> n;
    vector ans(n,vector<int>(n));
    int cnt = 1;
    for(int i = 0;i <= n / 2;i ++){
        int j = i, k = i + 1;
        while(j < n && !ans[i][j]) ans[i][j ++] = cnt ++;
        j --;
        while(k < n && !ans[k][j]) ans[k ++][j] = cnt ++;
        k --,j --;
        while(j >= 0 && !ans[k][j]) ans[k][j--] = cnt ++;
        j ++,k --;
        while(k >= 0 && !ans[k][j]) ans[k --][j] = cnt ++;
    }

    for(int i = 0;i < n;i ++){
        for(int j = 0;j < n;j ++){
            int num = to_string(ans[i][j]).size();
            for(int k = 0;k < 3 - num;k ++)
                cout << ' ';
            cout << ans[i][j];
        }
        cout << '\n';
    }
    return 0;
}

7-8 抓老鼠啊~亏了还是赚了？ - SMU 2024 spring 天梯赛4 (pintia.cn)

f1为不高兴期,f2为伤心期,g为兴奋期,其余按题意模拟;

#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 ans = 0;
    int f1 = 0, f2 = 0, g = 0;
    for (int i = 0; i < s.size() - 1; i ++) {
        if (!g) {
            if (f1 || f2) {
                if (f1) f1 --;
                if (f2) f2 --;
                cout << '-';
                continue;
            }
        }
        if(f1) f1 --;
        if(f2) f2 --;
        if(g) g --;
        if (s[i] == 'X') {
            f1 = 1;
            cout << 'U';
        } else if (s[i] == 'T') {
            f2 = 2;
            ans += 7;
            cout << 'D';
        } else if (s[i] == 'C') {
            f1 = f2 = 0;
            g = 2;
            ans -= 3;
            cout << '!';
        }
    }
    cout <<  '\n' << ans << '\n';

    return 0;
}

7-9 Windows消息队列 - SMU 2024 spring 天梯赛4 (pintia.cn)

用优先队列按优先级为关键字模拟;

#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;
    using a2 = pair<int,string>;
    priority_queue<a2,vector<a2>,greater<a2>> Q;
    while(n --){
        string s;
        cin >> s;
        if(s == "PUT"){
            string a;
            int x;
            cin >> a >> x;
            Q.push({x,a});
        }else{
            if(Q.empty()){
                cout << "EMPTY QUEUE!\n";
                continue;
            }
            cout << Q.top().second << '\n';
            Q.pop();
        }
    }

    return 0;
}

7-10 名人堂与代金券 - SMU 2024 spring 天梯赛4 (pintia.cn)

模拟;

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

struct Node {
    int x;
    string s;
    bool operator < (const Node& y) const {
        if (x != y.x) return x > y.x;
        return s < y.s;
    }
};

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

    int n, g, k;
    cin >> n >> g >> k;
    string s;
    vector<Node> res(n);

    int ans = 0;
    for (int i = 0; i < n; i ++) {
        int x;
        cin >> s >> x;
        res[i] = {x, s};
        if (x >= g) ans += 50;
        else if (x >= 60)  ans += 20;
    }

    cout << ans << '\n';

    sort(res.begin(), res.end());
    int cnt = 1, las = 0, now = 0;
    while (cnt <= k) {
        las = now;
        cout << cnt << ' ' << res[now].s << ' ' << res[now].x << '\n';
        now ++;
        while (now < n && res[now].x == res[now - 1].x) {
            cout << cnt << ' ' << res[now].s << ' ' << res[now].x << '\n';
            now ++;
        }
        cnt += now - las;
    }

    return 0;
}

7-11 用扑克牌计算24点 - SMU 2024 spring 天梯赛4(补题) (pintia.cn)

括号有三种类型,一种是2 3 12 12,((3-2)*12)+12,一种是2 13 7 7, 2*(13-(7/7)),还有一种是5 5 5 5,(5*5)-(5/5),注意判断的时候即可,其余可以直接用全排列然后去枚举四种符号即可;

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

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

    double a[4] = {};

    for (int i = 0; i < 4; i ++)
        cin >> a[i];

    auto cal = [](double x, double y, char c) -> double {
        if (c == '+') return x + y;
        if (c == '-') return x - y;
        if (c == '*') return x * y;
        return x / y;
    };
    sort(a, a + 4);
    char c[] = {'+', '-', '*', '/'};
    do {
        for (int i = 0; i < 4; i ++) {
            for (int j = 0; j < 4; j ++) {
                for (int k = 0; k < 4; k ++) {
                    if (cal(cal(cal(a[0], a[1], c[i]), a[2], c[j]), a[3], c[k]) == 24) {
                        cout << "((" << a[0] << c[i] << a[1] << ')' << c[j] << a[2] << ')' << c[k] << a[3];
                        return 0;
                    } else if (cal(cal(a[0], a[1], c[i]), cal(a[2], a[3], c[j]) , c[k]) == 24) {
                        cout << "(" << a[0] << c[i] << a[1] << ")" << c[k] << "(" << a[2] << c[j] << a[3] << ")";
                        return 0;
                    } else if (cal(a[0], cal(a[1], cal(a[2], a[3], c[k]), c[j]), c[i]) == 24) {
                        cout << a[0] << c[i] << "(" << a[1] << c[j] << "(" << a[2] << c[k] << a[3] << "))" ;
                        return 0;
                    }
                }
            }
        }
    } while (next_permutation(a, a + 4));

    cout << "-1\n";

    return 0;
}

7-12 玩转二叉树 - SMU 2024 spring 天梯赛4(补题) (pintia.cn)

注意节点没有说是\(1\sim n\)啊QAQ

写得时候一直段错误,后来改成存下标就过了,如果不理解这题的可以先去看看已知前序和中序怎么求后序,在递归的时候其实就是按左右儿子去递归,所以我们存的时候可以先存右儿子再存左儿子,这样就可以保证遍历的时候是反转的;

#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 + 1), mid(n + 1);
	vector g(n + 1, vector<int>());
	for (int i = 1; i <= n; i ++) cin >> mid[i];
	for (int i = 1; i <= n; i ++) cin >> pre[i];

	auto dfs = [&](auto dfs, int root, int start, int end) -> void{
		if (start > end)
			return ;
		int i = start;
		while (i < end && mid[i] != pre[root])
			i++;
		int l = root + 1, r = root + 1 - start + i;

		if (r <= n && r > 0)
			g[root].push_back(r);
		if (l > 0 && l <= n)
			g[root].push_back(l);

		dfs(dfs, r, i + 1, end); 

		dfs(dfs, l, start, i - 1);

	};

	dfs(dfs, 1, 1, n);

	queue<int> Q;
	Q.push(1);
	vector<bool> vis(n + 1);
	vector<int> ans;
	while (Q.size()) {
		int u = Q.front();
		Q.pop();
		if (vis[u]) continue;
		vis[u] = true;
		ans.push_back(pre[u]);
		for (auto v : g[u]) {
			Q.push(v);
		}
	}

	for (auto i : ans)
		cout << i << " \n"[i == ans.back()];

	return 0;
}

7-13 六度空间 - SMU 2024 spring 天梯赛4(补题) (pintia.cn)

bfs的进阶一点小技巧(,就是记录每层的最后一个节点last,然后当该次bfs遍历到了这一层的最后一个节点,那么久层数level++,大于6的时候退出即可;

dfs的话大多数题解也讲了,就是会丢掉某些最优解导致不能AC;

#include <bits/stdc++.h>

using namespace std;

using i64 = long long ;

int main() {

	int n, m;
	cin >> n >> m;
	vector g(n + 1, vector<int>());
	vector<bool> vis(n + 1);
	for (int i = 0; i < m; i ++) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
	}

	auto bfs = [&](int x) {
		vector<bool>(n + 1).swap(vis);
		int level = 0, last = x, now = 0, ans = 1;
		queue<int> Q;
		Q.push(x);
		vis[x] = 1;
		while (Q.size()) {
			auto u = Q.front();
			Q.pop();

			for (auto v : g[u]) {
				if (!vis[v]) {
					ans ++;
					Q.push(v);
					now = v;
					vis[v] = 1;
				}
			}

			if (u == last) {
				level ++;
				last = now;
			}

			if (level >= 6) {
				break;
			}
		}

		return ans;
	};

	for (int i = 1; i <= n; i ++) {
		int ans = bfs(i);
		printf("%d: %.2lf%%\n", i, ans * 100.0 / n);
	}

	return 0;
}