Solution Set -「ABC 205」

应该是最近最水的 ABC 了吧。

「ABC 205A」kcal

Link.

略

#include <bits/stdc++.h>
using ll = long long;
#define all(x) (x).begin(), (x).end()
int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
	ll a, b;
	std::cin >> a >> b;
	std::cout << b * a / 100.0 << "
";
	return 0;
}

「ABC 205B」Permutation Check

Link.

排序 / std::set 均可。

#include <bits/stdc++.h>
using ll = long long;
#define all(x) (x).begin(), (x).end()
int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
	int n, cur = 0;
	std::cin >> n;
	std::vector<int> a(n);
	for (int &x : a) {
		std::cin >> x;
		--x;
	}
	std::sort(all(a));
	for (int x : a) {
		if (cur != x) {
			std::cout << "No
";
			return 0;
		}
		++cur;
	}
	std::cout << "Yes
";
	return 0;
}

「ABC 205C」POW

Link.

若 (c) 为偶数则 (a:=|a|,b:=|b|)，然后比较 (a,b) 大小即可。

#include <bits/stdc++.h>
using ll = long long;
#define all(x) (x).begin(), (x).end()
int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
	int a, b, c;
	std::cin >> a >> b >> c;
	if (c % 2 == 0) {
		a = std::abs(a);
		b = std::abs(b);
	}
	if (a > b) std::cout << ">
";
	else if (a < b) std::cout << "<
";
	else std::cout << "=
";
	return 0;
}

「ABC 205D」Kth Excluded

Link.

预处理每一个数空出来的位置，然后询问时二分分类讨论。

#include <bits/stdc++.h>
using ll = long long;
#define all(x) (x).begin(), (x).end()
int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
	int n, q;
	std::cin >> n >> q;
	std::vector<ll> a(n), b(n);
	for (ll &x : a) std::cin >> x;
	for (size_t i = 0; i < a.size(); ++i) b[i] = a[i] - i - 1;
	for (ll k; q; --q) {
		std::cin >> k;
		ll pos = std::lower_bound(all(b), k) - b.begin();
		if (pos == n) std::cout << a.back() + k - b.back() << "
";
		else std::cout << a[pos] - b[pos] + k - 1 << "
";
	}
	return 0;
}

「ABC 205E」White and Black Balls

Link.

答案显然是 (inom{n+m}{n}-inom{n+m}{n-k-1})。

#include <bits/stdc++.h>
using ll = long long;
#define all(x) (x).begin(), (x).end()
int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
	constexpr int MOD = 1e9 + 7;
	int n, m, k;
	std::cin >> n >> m >> k;
	std::vector<ll> fac(n + m + 1), ifac(n + m + 1);
	auto pow = [&] (ll x, int y) {
		ll res = 1;
		for (; y; y >>= 1, x = x * x % MOD)
			if (y & 1) res = res * x % MOD;
		return (res + MOD) % MOD;
	};
	fac[0] = ifac[0] = 1;
	for (int i = 1; i < n + m + 1; ++i) {
		fac[i] = fac[i - 1] * i % MOD;
		ifac[i] = pow(fac[i], MOD - 2);
	}
	auto C = [&] (int n, int k) {return n < k ? 0 : fac[n] * ifac[n - k] % MOD * ifac[k] % MOD;};
	if (n - m > k) std::cout << "0
";
	else std::cout << (C(n + m, n) - C(n + m, n - k - 1) + MOD) % MOD << "
"; 
	return 0;
}

「ABC 205F」Grid and Tokens

Link.

网络流板题。

#include <bits/stdc++.h>
#include <atcoder/maxflow>
using ll = long long;
#define all(x) (x).begin(), (x).end()
int main() {
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
	int h, w, n;
	std::cin >> h >> w >> n;
	std::vector<std::vector<int>> obj(n, std::vector<int>(2));
	std::vector<int> row(h), col(w);
	auto id = [&] () {
		static int cnt = 0;
		return cnt++;
	};
	const int S = id(), T = id();
	for (int &x : row) x = id();
	for (int &x : col) x = id();
	for (std::vector<int> &x : obj) x = std::vector<int>({id(), id()});
	atcoder::mf_graph<int> G(id());
	for (int x : row) G.add_edge(S, x, 1);
	for (int x : col) G.add_edge(x, T, 1);
	for (int i = 0; i < n; ++i) {
		int a, b, c, d;
		std::cin >> a >> b >> c >> d;
		--a, --b;
		G.add_edge(obj[i][0], obj[i][1], 1);
		for (int j = a; j < c; ++j) G.add_edge(row[j], obj[i][0], 1);
		for (int j = b; j < d; ++j) G.add_edge(obj[i][1], col[j], 1);
	}
	std::cout << G.flow(S, T) << "
";
	return 0;
}