AtCoder Beginner Contest 366 题解：

\(Problem \hspace{2mm} A - Election \hspace{2mm} 2\)

题目链接

opinion:

很显然，当一个人的票数大于等于 \(\lceil \frac{n}{2} \rceil\) 时此人一定当选。（或可理解为投票结果一定固定。）

依次判断两人即可。

code:

#include <bits/stdc++.h>
#define ll long long
#define pii pair <int, int>
using namespace std;
const int N = 1e6 + 10;
int n, t, a;
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin >> n >> t >> a;
	int k = (n + 1) / 2;
	if (t >= k || a >= k) cout << "Yes\n";
	else cout << "No\n";
	return 0;
}

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\(Problem \hspace{2mm} B - Vertical \hspace{2mm} Writing\)

题目链接

opinion:

这题，呃，我竟然吃了发罚时……

试着把样例顺时针旋转 \(90^\circ\)，是不是与输出几乎一致？

为了方便，在每一个空位填上字符 *，又由于每个串结尾不能有 *，再将末尾连续的 * 去掉即可。

详见代码。

也就是按照题意模拟一下。

code:

#include <bits/stdc++.h>
#define ll long long
#define pii pair <int, int>
using namespace std;
const int N = 1e6 + 10;
int n, len[N];
string s[N], t[N];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin >> n;
	int m = 0;
	for (int i = 1; i <= n; ++i) {
		cin >> s[i];
		len[i] = s[i].size();
		s[i] = " " + s[i];
		m = max(m, len[i]);
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= len[i]; ++j) {
			t[j] = t[j] + s[i][j]; 
		}
		//前面有字符的就直接copy
		for (int j = len[i] + 1; j <= m; ++j) {
			t[j] = t[j] + '*';
		}
		//否则加 '*'
	}
	for (int i = 1; i <= m; ++i) {
		reverse(t[i].begin(), t[i].end());
		int l = t[i].size();
		int j = l - 1;
		while (t[i][j] == '*') j --;
		//找出末尾连续的 '*'，并将末尾更改
		for (int k = 0; k <= j; ++k)
			cout << t[i][k];
		cout << "\n";
	}
	return 0;
}

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\(Problem \hspace{2mm} C - Balls \hspace{2mm} and \hspace{2mm} Bag \hspace{2mm} Query\)

题目链接

opinion:

code:

#include <bits/stdc++.h>
#define ll long long
#define pii pair <int, int>
using namespace std;
const int N = 1e6 + 10;
int q, cnt[N];
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin >> q;
	int ans = 0;
	while (q -- ) {
		int op, x;
		cin >> op;
		if (op == 1) {
			cin >> x;
			cnt[x] ++;
			if (cnt[x] == 1) ans ++;
		} else if (op == 2) {
			cin >> x;
			cnt[x] --;
			if (cnt[x] == 0) ans --;
		} else {
			cout << ans << "\n";
		}
	}
	return 0;
}

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\(Problem \hspace{2mm} D - Reachability \hspace{2mm} in \hspace{2mm} Functional \hspace{2mm} Graph\)

题目链接

opinion:

code:

#include <bits/stdc++.h>
#define ll long long
#define pii pair <int, int>
using namespace std;
const int N = 110;
ll a[N][N][N];
ll pre[N][N][N];
inline ll query(int N1, int M1, int K1, int N2, int M2, int K2) {
    return pre[N2][M2][K2] - pre[N1 - 1][M2][K2] - pre[N2][M1 - 1][K2] - pre[N2][M2][K1 - 1] + pre[N1 - 1][M1 - 1][K2] + pre[N1 - 1][M2][K1 - 1] + pre[N2][M1 - 1][K1 - 1] - pre[N1 - 1][M1 - 1][K1 - 1];
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n; cin >> n;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			for (int k = 1; k <= n; ++k) {
				cin >> a[i][j][k];
				pre[i][j][k] = a[i][j][k] + pre[i - 1][j][k] + pre[i][j - 1][k] + pre[i][j][k - 1] - pre[i - 1][j - 1][k] - pre[i - 1][j][k - 1] - pre[i][j - 1][k - 1] + pre[i - 1][j - 1][k - 1];
			}
		}
	}
	int q; cin >> q;
	while (q -- ) {
		ll l[3], r[3];
		cin >> l[0] >> r[0] >> l[1] >> r[1] >> l[2] >> r[2];
		ll ans = query(l[0], l[1], l[2], r[0], r[1], r[2]);
		cout << ans << "\n";
	}
	return 0;
}

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