Codeforces Round #875 (Div. 2)

Codeforces Round #875 (Div. 2)

bilibili: Codeforces Round #875 (Div. 2) 实况 | 完成度 [4.01 / 6]

A

#include <bits/stdc++.h>
using ll = long long;
using uint = unsigned;
using namespace std;

int TestCase;

signed main() {
	for (scanf("%d", &TestCase); TestCase--;) {
		int n;
		scanf("%d", &n);
		vector<int> a(n);
		for (int &x : a) scanf("%d", &x);
		for (int i = 0; i < n; ++i) {
			printf("%d%c", n - a[i] + 1, " \n"[i == n - 1]);
		}
	}
	return 0;
}

B

#include <bits/stdc++.h>
using ll = long long;
using uint = unsigned;
using namespace std;

int TestCase;

signed main() {
	for (scanf("%d", &TestCase); TestCase--;) {
		int n;
		scanf("%d", &n);
		vector<int> a(n), b(n);
		for (int &x : a) scanf("%d", &x);
		for (int &x : b) scanf("%d", &x);
		vector<int> t(2 * n + 100, 0);
		int ans = 0;
		for (int len = 0, i = 0; i < n; ++i) {
			++len;
			if (i == 0 || a[i] != a[i - 1]) len = 1;
			ans = max(ans, len);
			t[a[i]] = max(t[a[i]], len);
		}
		for (int len = 0, i = 0; i < n; ++i) {
			++len;
			if (i == 0 || b[i] != b[i - 1]) len = 1;
			// ans = max(ans, len);
			ans = max(ans, len + t[b[i]]);
		}
		printf("%d\n", ans);
	}
	return 0;
}

C（1A）

#include <bits/stdc++.h>
using ll = long long;
using uint = unsigned;
using namespace std;

int TestCase;

signed main() {
	for (scanf("%d", &TestCase); TestCase--;) {
		int n;
		scanf("%d", &n);
		vector<vector<pair<int, int>>> G(n);
		vector<int> drawn(n, 0), mp(n, 0);
		for (int i = 1; i < n; ++i) {
			int u, v;
			scanf("%d%d", &u, &v);
			--u, --v;
			G[u].emplace_back(v, i);
			G[v].emplace_back(u, i);
		}
		set<int> all;
		drawn[0] = 1;
		function<void(int)> access = [&](int u) {
			for (auto [v, id] : G[u])
				if (!drawn[v]) all.emplace(id), mp[id] = v;
		};
		int total = 1;
		access(0);
		for (int cur = 0; all.size(); ) {
			auto it = all.lower_bound(cur);
			if (it == end(all)) it = all.lower_bound(0), ++total;
			cur = *it;
			drawn[mp[cur]] = 1;
			access(mp[cur]);
			all.erase(cur);
		}
		printf("%d\n", total);
	}
	return 0;
}

D（1B）

#include <bits/stdc++.h>
using ll = long long;
using uint = unsigned;
using namespace std;

int TestCase;

signed main() {
	for (scanf("%d", &TestCase); TestCase--;) {
		int n;
		scanf("%d", &n);
		vector<int> a(n), b(n);
		for (int &x : a) scanf("%d", &x);
		for (int &x : b) scanf("%d", &x);
		// a[i] * a[j] = b[i] + b[j] <= 2 * n
		const int top = 2 * n;
		vector<unordered_map<int, ll>> all(n * 2 + 10);
		for (int i = 0; i < n; ++i) ++all[a[i]][b[i]];
		ll ans = 0;
		for (int x = 1; x * x <= top && x <= n; ++x)
			for (int y = x + 1; y <= n && x * y <= top; ++y) {
				int val = x * y;
				// printf("x = %d, y = %d\n", x, y);
				if (all[x].size() < all[y].size()) {
					for (auto [v1, cnt] : all[x])
						if (all[y].find(val - v1) != end(all[y])) ans += (ll)cnt * all[y][val - v1];
				} else {
					for (auto [v1, cnt] : all[y])
						if (all[x].find(val - v1) != end(all[x])) ans += (ll)cnt * all[x][val - v1];
				}
				// printf("ans = %lld\n", ans);
			}
		for (int x = 1; x * x <= top && x <= n; ++x) {
			int val = x * x;
			for (auto [v, cnt] : all[x])
				if (v * 2 == val) {
					ans += (ll)cnt * (cnt - 1) / 2;
				} else if (v < val - v && all[x].find(val - v) != end(all[x])) ans += (ll)cnt * all[x][val - v];
			// printf("x = %d, ans = %lld\n", x, ans);
		}
		printf("%lld\n", ans);
	}
	return 0;
}
// ？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？

E（1C）（补）

#include <bits/stdc++.h>
using ll = long long;
using uint = unsigned;
using ull = unsigned long long;
using namespace std;

constexpr int mod = 998244353;
// assume -mod <= x < 2mod
int normZ(int x) {
	if (x < 0) x += mod;
	if (x >= mod) x -= mod;
	return x;
}
template <typename T> T qpow(T x, ll k) {
	T res = 1;
	for (; k; k >>= 1, x *= x)
		if (k & 1) res *= x;
	return res;
}

struct Z {
	int x;
	Z(int x = 0) : x(normZ(x)) {}
	Z(ll x) : x(normZ(x % mod)) {}
	int val() const { return x; }
	Z operator-() const { return Z(normZ(mod - x)); }
	Z inv() const {
		assert(x != 0);
		return qpow(*this, mod - 2);
	}
	Z &operator*=(const Z &rhs) {
		x = (ll)x * rhs.x % mod;
		return *this;
	}
	Z &operator+=(const Z &rhs) {
		x = normZ(x + rhs.x);
		return *this;
	}
	Z &operator-=(const Z &rhs) {
		x = normZ(x - rhs.x);
		return *this;
	}
	Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
	friend Z operator*(const Z &lhs, const Z &rhs) {
		Z res = lhs;
		res *= rhs;
		return res;
	}
	friend Z operator+(const Z &lhs, const Z &rhs) {
		Z res = lhs;
		res += rhs;
		return res;
	}
	friend Z operator-(const Z &lhs, const Z &rhs) {
		Z res = lhs;
		res -= rhs;
		return res;
	}
	friend Z operator/(const Z &lhs, const Z &rhs) {
		Z res = lhs;
		res /= rhs;
		return res;
	}
};

const int mxn = 6e5 + 10;

Z fct[mxn], inv[mxn], ift[mxn];

Z C(int n, int m) {
	if (n < m || m < 0) return 0;
	return fct[n] * ift[m] * ift[n - m];
}

int TestCase, n, k;

// [l1, r1], [l2, r2]
// => [l1, l2), [l2, r1], (r1, r2]
// < ( > )
// 对于相交的区间，可以拆分成三个区间，如上
// 最后会得到一堆不相交但包含的线段
// dp 上去，时间复杂度总线段数，基本上不超过 2n, 4n
// <  ---- ----   ----> f[cur] = G(empty) * f[sub interval], G(n) 长度为 n 的括号序列的方案数

Z Cat(int n) {
	return C(2 * n, n) * inv[n + 1];
}

ull a[mxn];

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

signed main() {
	fct[0] = inv[0] = ift[0] = fct[1] = inv[1] = ift[1] = 1;
	for (int i = 2; i <= 6e5 + 3; ++i) {
		fct[i] = fct[i - 1] * i;
		inv[i] = inv[mod % i] * (mod - mod / i);
		ift[i] = ift[i - 1] * inv[i];
	}
	for (scanf("%d", &TestCase); TestCase--;) {
		scanf("%d%d", &n, &k);
		fill(a, a + n + 1, 0);
		for (int i = 1; i <= k; ++i) {
			int l, r; scanf("%d%d", &l, &r);
			ull x = rng();
			a[l] ^= x, a[r + 1] ^= x;
		}
		map<ull, int> cnt;
		for (int i = 1; i <= n; ++i) {
			a[i] ^= a[i - 1];
			++cnt[a[i]];
		}
		Z ans = 1;
		for (auto [_, v] : cnt) ans *= (v & 1 ? Z(0) : Cat(v / 2));
		printf("%d\n", ans.val());
	}
	return 0;
}