Codeforces Round 966 (Div. 3) VP

A. Primary Task

列舉所有情況即可

void solve(){
	string s;
	cin >> s;

	if (s.substr(0, 2) != "10" || s.size() < 3 || s[2] == '0' || (s.size() < 4 && s[2] < '2')) {
		cout << "NO\n";
	}
	else {
		cout << "YES\n";
	}
}

B. Seating in a Bus

簡單模擬題

void solve(){
	int n;
	cin >> n;

	vector<int> a(n + 2);
	bool ok = true;
	for (int i = 1; i <= n; ++i) {
		int t;
		cin >> t;
		if (!ok) {
			continue;
		}
		if (i == 1) {
			a[t] = 1;
		}
		else {
			if (a[t - 1] || a[t + 1]) {
				a[t] = 1;
			}
			else {
				ok = false;
			}
		}
	}

	cout << (ok ? "YES\n" : "NO\n");
}

C. Numeric String Template
一時沒想到低空間複雜度解法，直接上了map+並查集

void solve(){
	int n;
	cin >> n;

	vector<int> a(n);
	for (auto& x : a) {
		cin >> x;
	}

	int m;
	cin >> m;

	bool ok = false;
	map<int, vector<int>> mapp;

	for (int i = 0; i < n; ++i) {
		mapp[a[i]].push_back(i);
	}

	Dsu dsu(n + 1);
	for (const auto&[x, y] : mapp) {
		for (int i = 0; i < (int)y.size() - 1; ++i) {
			dsu.unionSet(y[i], y[i + 1]);
		}
	}

	for (int i = 0; i < m; ++i) {
		string s;
		cin >> s;

		if (s.size() != n) {
			cout << "NO\n";
			continue;
		}
		mapp.clear();
		for (int i = 0; i < n; ++i) {
			mapp[s[i]].push_back(i);
		}
		Dsu dsu2(n + 1);
		for (const auto&[x, y] : mapp) {
			for (int i = 0; i < (int)y.size() - 1; ++i) {
				dsu2.unionSet(y[i], y[i + 1]);
			}
		}

		bool same = true;
		for (int i = 0; i < n; ++i) {
			if (dsu.findSet(i) != dsu2.findSet(i)) {
				same = false;
				break;
			}
		}
		cout << (same ? "YES" : "NO") << '\n';
	}

}

D. Right Left Wrong

貪心問題，策略應該是讓數字儘可能被多次數的操作，於是進行LR匹配即可

void solve(){
	int n;
	cin >> n;

	vector<int> a(n);
	for (auto& x : a) {
		cin >> x;
	}

	string s;
	cin >> s;

	long long ans = 0;
	vector<pair<int, int>> pairs;

	int l = -1;
	int r = n;
	while (l <= r) {
		size_t ll = s.find_first_of('L', l + 1);
		size_t rr = s.find_last_of('R', r - 1);
		if ((ll == s.npos) || (rr == s.npos)) {
			break;
		}
		if (ll <= rr) {
			pairs.emplace_back(ll, rr);
		}
		l = ll;
		r = rr;
	}

	vector<long long> pref(n + 1);
	for (int i = 0; i < n; ++i) {
		pref[i + 1] = pref[i] + a[i];
	}


	for (const auto&[x, y] : pairs) {
		ans += (pref[y + 1] - pref[x]);
	}

	cout << ans << '\n';

}

E. Photoshoot for Gorillas

跟D的出題意願差不多，依然是統計出現次數最多的格子，不過是在二維中
二維差分即可
這裡做的時候卡住了，第一次寫二維差分，多減了一次權值沒有補回來

void solve(){
	int n, m, k;
	cin >> n >> m >> k;

	int q;
	cin >> q;
	vector<int> a(q);
	for (auto& x : a) {
		cin >> x;
	}

	vector<vector<int>> grid(n + 2, vector<int>(m + 2));
	for (int i = 1; i + k <= n + 1; ++i) {
		for (int j = 1; j + k <= m + 1; ++j) {
			grid[i][j] ++;
			grid[i][j + k] --;
			grid[i + k][j] --;
			grid[i + k][j + k] ++;
			//for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) cout << grid[i][j] << " \n"[j == m];
			//cout << endl;
		}
	}

	multiset<int> sett;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			grid[i][j] += (grid[i][j - 1] + grid[i - 1][j] - grid[i - 1][j - 1]);
			int t = grid[i][j];
			sett.insert(grid[i][j]);
			//for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) cout << grid[i][j] << " \n"[j == m];
			//cout << endl;
		}
	}

	long long ans = 0;
	sort(a.rbegin(), a.rend());

	auto it = sett.rbegin();
	for (int i = 0; i < q; ++i) {
		ans += 1ll * a[i] * (*it);
		it ++;
		if (it == sett.rend()) {
			break;
		}
	}

	cout << ans << '\n';

}