A. Robin Helps
顺着题意做一遍模拟即可，代码如下：

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#include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;

int main() {
	int t;
	cin >> t;
	while (t--) {
		int n, k;
		cin >> n >> k;
		ll temp = 0;
		ll ans = 0;
		vector<ll> a(n);
		for (int i = 0; i < n; i++) {
			cin >> a[i];
			if (a[i] >= k) {
				temp += a[i];
			}
			if (a[i] == 0) {
				if (temp > 0) {
					ans++;
					temp--;
				}
			}
		}
		cout << ans << endl;
	}
	return 0;
}

B. Robin Hood and the Major Oak
此题硬写会超时，判断奇偶性即可，代码如下：

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#include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;
const ll MOD = 1e9 + 5;
int t;

int main() {
	cin >> t;
	while (t--) {
		ll cnt = 0;
		ll n, k;
		cin >> n >> k;
		if (n % 2 == 0) {
			if ((k / 2) % 2 == 0) {
				cout << "YES" << endl;
			} else {
				cout << "NO" << endl;
			}
		} else {
			if ((ll)ceil(k * 1.0 / 2) % 2 == 0) {
				cout << "YES" << endl;
			} else {
				cout << "NO" << endl;
			}
		}
	}
	return 0;
}

C. Robin Hood in Town
先输入钱数并进行排序，找到满足题意的临界点的那个人，将他的钱数 * 2 * n得到满足题目的条件的sum,然后将以输入的实际钱数求和，与sum作比较，得出答案。(需要注意的是，当n == 1 || n == 2时, 需要输出-1, 此为特判)

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#include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		vector<ll> a(n);
		ll sum2 = 0;
		for (int i = 0; i < n; i++) {
			cin >> a[i];
			sum2 += a[i];
		}
		if (n <= 2) {
			cout << -1 << endl;
			continue;
		}
		sort(a.begin(), a.end());
		ll mid = n / 2;
		ll sum1 = a[mid] * 2 * n + 1;
		if (sum1 <= sum2) {
			cout << 0 << endl;
		} else {
			cout << sum1 - sum2 << endl;
		}
	}
	return 0;
}

D. Robert Hood and Mrs Hood
使用差分数组来计算区间贡献，但需要注意的是，访客探访的时间是d，因此在计算贡献时，应计算l - d + 1。

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#include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;
vector<ll> diff(200005);

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int t;
	cin >> t;
	while (t--) {
		int n, d, k;
		cin >> n >> d >> k;
		for (int i = 1; i <= k; i++) {
			ll l, r;
			cin >> l >> r;
			l = max(1ll, l - d + 1);
			diff[l]++;
			diff[r + 1]--;
		}
		for (int i = 1; i <= n; i++) {
			diff[i] += diff[i - 1];
		}
		ll maxx = -1, minn = 1e10;
		ll mum, bro;
		for (int i = 1; i <= n - d + 1; i++) {
			if (diff[i] > maxx) {
				maxx = diff[i];
				bro = i;
			}
			if (diff[i] < minn) {
				minn = diff[i];
				mum = i;
			}
		}
		cout << bro << " " << mum << endl;
		for (int i = 0; i <= n + 1; i++) {
		diff[i] = 0;
	}
	}
	return 0;
}