HHKB Programming Contest 2020【ABCE】

比赛链接：https://atcoder.jp/contests/hhkb2020/tasks

A - Keyboard

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	char s, t;
	cin >> s >> t;
	cout << (s == 'Y' ? char(toupper(t)) : t) << "\n";
	return 0;
}

B - Futon

题解

每个点只考虑右方和下方的点即可。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int h, w;
	cin >> h >> w;
	vector<string> MP(h);
	for (auto &x : MP) cin >> x;
	int ans = 0;
	for (int i = 0; i < h; i++) {
		for (int j = 0; j < w; j++) {
			if (MP[i][j] == '.' and i + 1 < h and MP[i + 1][j] == '.') ++ans;
			if (MP[i][j] == '.' and j + 1 < w and MP[i][j + 1] == '.') ++ans;
		}
	}
	cout << ans << "\n";
	return 0;
}

C - Neq Min

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n;
	cin >> n;
	set<int> st;
	for (int i = 0; i <= 200010; i++)
		st.insert(i);
	for (int i = 0; i < n; i++) {
		int x;
		cin >> x;
		st.erase(x);
		cout << *st.begin() << "\n";
	}
	return 0;
}

E - Lamps

题解

假设每盏灯在所有情况中都亮着，则亮着的灯的总数为 \(k \cdot 2^k\) 。
考虑每盏灯不亮的情况有多少种：一盏灯不亮的充要条件是上下左右连通的灯都不亮，设这些灯加上自身总个数为 \(tot\)，那么其余的 \(k-tot\) 盏灯的亮灭情况是随意的，即 \(2^{(k - tot)}\) 。
答案即为 $k \cdot 2^k - \sum \limits _{i = 1}^k 2^{(k - tot_i)} $ 。
上下左右连通的灯数用前缀和计算一下即可。

代码

#include <bits/stdc++.h>
#define int long long
using namespace std;
constexpr int N = 2020;
constexpr int MOD = 1e9 + 7;

char MP[N][N];
int up[N][N];
int dn[N][N];
int lf[N][N];
int rt[N][N];
int k;

int binpow(int a, int b) {
	int res = 1;
	while (b) {
		if (b & 1) res = 1LL * res * a % MOD;
		a = 1LL * a * a % MOD;
		b >>= 1;
	}
	return res;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int h, w;
	cin >> h >> w;
	for (int i = 1; i <= h; i++) {
		for (int j = 1; j <= w; j++) {
			cin >> MP[i][j];
			if (MP[i][j] == '.') ++k;
		}
	}
	for (int j = 1; j <= w; j++) {
		for (int i = 1; i <= h; i++) {
			if (MP[i][j] == '#') {
				up[i][j] = 0;
			} else {
				up[i][j] = up[i - 1][j] + 1;
			}
		}
	}
	for (int j = 1; j <= w; j++) {
		for (int i = h; i >= 1; i--) {
			if (MP[i][j] == '#') {
				dn[i][j] = 0;
			} else {
				dn[i][j] = dn[i + 1][j] + 1;
			}
		}
	}
	for (int i = 1; i <= h; i++) {
		for (int j = 1; j <= w; j++) {
			if (MP[i][j] == '#') {
				lf[i][j] = 0;
			} else {
				lf[i][j] = lf[i][j - 1] + 1;
			}
		}
	}
	for (int i = 1; i <= h; i++) {
		for (int j = w; j >= 1; j--) {
			if (MP[i][j] == '#') {
				rt[i][j] = 0;
			} else {
				rt[i][j] = rt[i][j + 1] + 1;
			}
		}
	}
	int ans = k * binpow(2, k);
	for (int i = 1; i <= h; i++) {
		for (int j = 1; j <= w; j++) {
			if (MP[i][j] == '.') {
				int tot = up[i][j] + dn[i][j] + lf[i][j] + rt[i][j] - 4 + 1;
				ans -= binpow(2, k - tot);
				(ans += MOD) %= MOD;
			}
		}
	}
	cout << ans << "\n";
	return 0;
}