周报

写在前言

放假到现在几乎没怎么训练与刷题,也不知道该怎么去描述,从回到家基本就待在医院陪老爸再到陪他最后一面,心情也很是复杂,其实我甚至想过退队,不知道该怎么说,也可能受到网上一些网友的言论影响,总之想多赚点钱,竞赛这条路从最开始的兴趣到现在开始迷茫了,老爸在的时候还想着兴趣在学校里也能打打算法竞赛觉得挺好的,现在只想把对老爸亏欠的都补偿在老妈身上,之前是陪老爸,现在是陪老妈,感觉已经没什么心思训练了,这次也是跟着打了一下训练赛,可能看上去还行,但是这些题看着其实也就是比较基础的模拟和一些基础算法,而这些对于自己来说题意和模拟也需要长时间去写,关于期望那题也没什么思路,觉得脑子挺乱的,烦死了,

补题及题解

SMU 2024 winter round 3

7-1 猫是液体

输出\(a \times b \times c\)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;

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

	i64 a,b,c;
	cin >> a >> b >> c;
	cout << a * b * c << '\n';

	return 0;
}

7-2 计算指数

输出\(2\)的\(n\)次方即可

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	i64 n, ans = 1;
	cin >> n;
	for (int i = 1; i <= n; i ++, ans *= 2);
	cout << "2^" << n << " = " << ans << '\n';

	return 0;
}

7-3 电子汪

输出\(a + b\)个\(Wang!\)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;

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

	int a, b;
	cin >> a >> b;
	for (int i = 0; i < a + b; i ++)
		cout << "Wang!";

	return 0;
}

7-4 最佳情侣身高差

男性除以\(1.09\),女性乘以\(1.09\)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;

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

	int n;
	cin >> n;
	while (n --) {
		char c;
		double a;
		cin >> c >> a;
		if (c == 'M') {
			printf("%.2lf\n", a / 1.09);
		} else {
			printf("%.2lf\n", a * 1.09);
		}
	}
	return 0;
}

7-5 不变初心数

对\(n\)的每一倍数挨个计算每一位的和,判断最后是否一样即可

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

void solve() {
	int n;
	cin >> n;
	int num = -1;
	for (int i = 2; i <= 9; i ++) {
		i64 x = i * n, sum = 0;
		while (x) {
			sum += x % 10;
			x /= 10;
		}
		if (num == -1) num = sum;
		else if (num != sum) {
			cout << "NO\n";
			return ;
		}
	}
	cout << num << '\n';
}

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

	int T;
	cin >> T;
	while (T --)
		solve();

	return 0;
}

7-6 吃火锅

暴力判断每句是否含有目标字符串即可,一句中含有多个目标按一个计算

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	string s, tar = "chi1 huo3 guo1";
	getline(cin, s);
	int ans = 0, first = 0, sum = 0;
	while (s != ".") {
		ans ++;
		for (int i = 0; i < s.size(); i ++) {
			if (s.substr(i, tar.size()) == tar) {
				if (!first) first = ans;
				sum ++;
				break;
			}
		}
		getline(cin, s);
	}
	if (!first) {
		cout << ans << '\n' << "-_-#\n";
	} else {
		cout << ans << '\n' << first << ' ' << sum << '\n';
	}
	return 0;
}

7-7 敲笨钟

把每句的拼音都单独存储,判断逗号和句号前面的是否都压了“ong”,是就替换最后三个拼音

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	int n;
	cin >> n;
	string s;
	getline(cin, s);
	while (n --) {
		getline(cin, s);
		vector<string> so;
		int kong = 0, i = 0, fi = 0, num = 0;
		for (; i < s.size(); i ++) {
			if (s[i] == ' ') {
				if (s[i - 1] == ',') fi = num;
				so.push_back(s.substr(kong, i - kong));
				kong = i + 1;
				num ++ ;
			}
		}
		so.push_back(s.substr(kong, i - kong));
		string op = so[fi];
		if (op.substr(max((int)op.size() - 4, 0), 3) == "ong" && so.back().substr(max((int)so.back().size(), 0) - 4, 3) == "ong") {
			for (int i = 0; i < so.size() - 3; i ++)
				cout << so[i] << ' ';
			cout << "qiao ben zhong.\n";
		} else
			cout << "Skipped\n";
	}

	return 0;
}

7-8 查找书籍

按价格排下序即可

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;
typedef pair<double, string> PSD;

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

	int n;
	cin >> n;
	vector<PSD> so(n);
	string s;
	getline(cin, s);
	for (int i = 0; i < n; i ++) {
		getline(cin, so[i].second);
		cin >> so[i].first;
		getline(cin, s);
	}

	sort(so.begin(), so.end());

	printf("%.2lf, %s\n", so.back().first, so.back().second.c_str());
	printf("%.2lf, %s\n", so[0].first, so[0].second.c_str());

	return 0;
}

7-9 梯云纵

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	int n;
	cin >> n;

	vector<double> a(110000), mid(110000);
	a[0] = 0, a[1] = 1, a[2] = 1.5, a[3] = 2.25;
	mid[0] = a[1] - a[0], mid[1] = a[2] - a[1], mid[2] = a[3] - a[2];

	auto getmid = [&](auto self, int x) -> double {
		return (mid[x] != 0 ? mid[x] : mid[x] = (self(self, x - 1) + self(self, x - 2)) / 2);
	};

	auto geta = [&](auto self, int x) -> double {
		return (a[x] != 0 ? a[x] : a[x] = (self(self, x - 1) + getmid(getmid, x - 1)));
	};

	while (n --) {
		int x;
		cin >> x;
		printf("%.6lf\n", geta(geta, x) + (x > 10000) * 0.0000002);
	}

	return 0;
}

7-10 家庭房产

用并查集将家庭联系起来,用家庭中编号最小的作为根节点,最后单独开个数组将家庭的各个数值计算出来,再排序即可.

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

template<typename T>
struct UFS {
	int sz;
	vector<T> rank, p;
	void link(T x, T y) {
		if (x == y)
			return;
		if (y > x)
			p[y] = x;
		else
			p[x] = y;
	}
	void init(int n) {
		sz = n;
		rank.resize(n + 1);
		p.resize(n + 1);
		for (int i = 0; i <= sz; i++) {
			p[i] = i;
			rank[i] = 0;
		}
	}
	T find(T x) {
		return x == p[x] ? x : (p[x] = find(p[x]));
	}
	void unin(T x, T y) {
		link(find(x), find(y));
	}
	void compress() {
		for (int i = 0; i < sz; i++)
			find(i);
	}
};

struct Home {
	int id, fa, ma, kid, all = 0;
	double fang = 0, S = 0;
	bool home = 0;
};

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

	int n;
	cin >> n;
	UFS<int> ufs;
	vector<Home> info(10086), ans(10086);
	vector<bool> vis(10086);
	ufs.init(10086);

	for (int i = 0; i < n; i ++) {
		int kid, k;
		cin >> info[i].id >> info[i].fa >> info[i].ma >> k ;
		vis[info[i].id] = 1;
		if (info[i].fa != -1) {
			ufs.unin(info[i].id, info[i].fa);
			vis[info[i].fa] = 1;
		}
		if (info[i].ma != -1) {
			ufs.unin(info[i].id, info[i].ma);
			vis[info[i].ma] = 1;
		}
		for (int j = 0; j < k; j ++) {
			cin >> kid;
			ufs.unin(kid, info[i].id);
			vis[kid] = 1;
		}
		cin >> info[i].fang >> info[i].S;
	}

	for (int i = 0; i < n; i ++) {
		int x = ufs.find(info[i].id);
		ans[x].id = x;
		ans[x].S += info[i].S;
		ans[x].fang += info[i].fang;
		ans[x].home = 1;
	}

	int num = 0;
	for (int i = 0; i < 10086; i ++) {
		if (vis[i]) ans[ufs.find(i)].all ++;
		if (ans[i].home) num ++;
	}

	for (int i = 0; i < 10086; i ++) {
		if (ans[i].home) {
			ans[i].S = ans[i].S / ans[i].all;
			ans[i].fang = ans[i].fang / ans[i].all;
		}
	}

	sort(ans.begin(), ans.end(), [&](Home a, Home b) {
		if (a.S != b.S) return a.S > b.S;
		return a.id < b.id;
	});

	printf("%d\n", num);
	for (int i = 0; i < num; i ++) {
		printf("%04d %d %.03lf %.03lf\n", ans[i].id, ans[i].all, ans[i].fang, ans[i].S);
	}

	return 0;
}

7-12 秀恩爱分得快

感觉本质上其实还是模拟,以及对题意的理解,

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

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

	const int N = 1010;
	vector<int> a(N), sex(N);
	vector g(N, vector<double>(N));
	while (m --) {
		int k, num = 0;
		cin >> k;
		for (int i = 0; i < k; i ++) {
			string s;
			cin >> s;
			int x;
			if (s[0] == '-') s[0] = '0', x = stoi(s), sex[x] = 1;
			else x = stoi(s);
			a[num ++] = x;
		}
		for (int i = 0; i < k - 1; i ++) {
			for (int j = i + 1; j < k; j ++) {
				if (sex[a[i]] == sex[a[j]]) continue;
				g[a[i]][a[j]] += 1.0 / k;
				g[a[j]][a[i]] = g[a[i]][a[j]];
			}
		}
	}

	string s;
	int x, y;
	cin >> s;
	if (s[0] == '-') s[0] = '0', x = stoi(s), sex[x] = 1;
	else x = stoi(s);
	cin >> s;
	if (s[0] == '-') s[0] = '0', y = stoi(s), sex[y] = 1;
	else y = stoi(s);

	double mx = 0, my = 0;
	for (int i = 0; i < n; i ++) {
		if (sex[i] != sex[x]) mx = max(mx, g[x][i]);
		if (sex[i] != sex[y]) my = max(my, g[y][i]);
	}

	if (mx == my && g[x][y] == mx) cout << (sex[x] ? "-" : "") << x << " " << (sex[y] ? "-" : "") << y << '\n';
	else {
		for (int i = 0; i < n; i ++)
			if (g[x][i] == mx && sex[x] != sex[i]) cout << (sex[x] ? "-" : "") << x << ' ' << (sex[i] ? "-" : "") << i << '\n';
		for (int i = 0; i < n; i ++)
			if (g[y][i] == my && sex[y] != sex[i]) cout << (sex[y] ? "-" : "") << y << ' ' << (sex[i] ? "-" : "") << i << '\n';

	}

	return 0;
}

7-13 地下迷宫探索

\(DFS\)从起点往下搜即可,判断到达的地方与\(n\)的关系,小于\(n\)则代表没有全部到达

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

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

	int n, m, s;
	cin >> n >> m >> s;
	vector g(n + 1, vector<int>());
	for (int i = 0; i < m; i ++) {
		int x, y;
		cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}

	for (int i = 1; i <= n; i ++) {
		sort(g[i].begin(), g[i].end());
	}

	vector<bool> vis(n + 1);
	int num = 0;
	auto dfs = [&](auto self, int x) -> void {

		if (num) cout << ' ';
		cout << x ;
		for (auto v : g[x]) {
			if (!vis[v]) {
				vis[v] = 1;
				num ++;
				self(self, v);
				cout << ' ' << x;
			}
		}

	};

	vis[s] = 1;
	dfs(dfs, s);
	if (++ num < n) cout << " 0\n";

	return 0;
}

7-14 畅通工程之最低成本建设问题

\(Kurskal\)算法跑一遍最小生成树即可,模板题吧

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

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

template<typename T>
struct UFS {
	T sz;
	vector<T> rank, p;
	void link(T x, T y) {
		if (x == y)
			return;
		if (rank[x] > rank[y])
			p[y] = x;
		else
			p[x] = y;
		if (rank[x] == rank[y])
			rank[y]++;
	}
	void init(int n) {
		sz = n;
		rank.resize(n + 1);
		p.resize(n + 1);
		for (int i = 0; i <= sz; i++) {
			p[i] = i;
			rank[i] = 0;
		}
	}
	T find(T x) {
		return x == p[x] ? x : (p[x] = find(p[x]));
	}
	void unin(T x, T y) {
		link(find(x), find(y));
	}
	void compress() {
		for (int i = 0; i < sz; i++)
			find(i);
	}
};

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

	int n, m;
	cin >> n >> m;
	vector<pair<int, pair<int, int>>> road(m);
	for (int i = 0; i < m; i ++) {
		cin >> road[i].second.first >> road[i].second.second;
		cin >> road[i].first;
	}

	sort(road.begin(), road.end());

	UFS<int> ufs;
	ufs.init(n);
	i64 ans = 0;
	for (int i = 0; i < m; i ++) {
		auto [x, y] = road[i].second;
		int dx = ufs.find(x), dy = ufs.find(y);
		if (dx != dy) {
			ufs.unin(dx, dy);
			ans += road[i].first;
		}
	}

	for (int i = 1; i <= n; i ++) {
		if (ufs.find(i) != ufs.find(1)) {
			cout << "Impossible\n";
			return 0;
		}
	}

	cout << ans << '\n';

	return 0;
}

posted @ 2024-02-19 00:01 Ke_scholar 阅读(16) 评论(0) 编辑收藏举报

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