Codeforces Round #642

A

分析可得：当$n\ge 3$时就能取得最大值$2m$

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	int t; cin >> t;
	while (t--) {
		int n, m; cin >> n >> m;
		if (n >= 3) cout << 2 * m << endl;
		else if (n == 2) cout << m << endl;
		else cout << 0 << endl;
	}
}

B

全部元素按大小排序后，能选k个b序列的就选，否则选a序列的。

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	
	int t; cin >> t;
	while (t--) {
		int n, k; cin >> n >> k;
		vector a(2 * n), vis(35);
		for (int i = 0; i < n; ++i) {
			cin >> a[i];
			vis[a[i]]++;
		}
		for (int i = 0; i < n; ++i) {
			cin >> a[i + n];
		}
		sort(a.begin(), a.begin() + 2 * n, [&](int x, int y) { return x > y; });
		int sum = 0, tmp_n = n;
		for (int i = 0; i < 2 * n && tmp_n; ++i) {
			if (vis[a[i]]) {
				sum += a[i];
				tmp_n--;
				vis[a[i]]--;
			}
			else if (k) {
				sum += a[i];
				tmp_n--;
				k--;
			}
		}
		cout << sum << endl;
	}
}

C

全部都要回到最中间，然后一圈一圈的发散，每圈的数字个数$8\ast i$，乘上该圈的步数$i$得到$\sum _{i=1}^{n/2}8\ast i^2$

int main()
{
	ios::sync_with_stdio(false); cin.tie(0);
	
	int t; cin >> t;
	while (t--) {
		int n; cin >> n;
		ll sum = 0;
		for (int i = 1; i <= (n - 1) / 2; ++i) {
			sum += 8ll * i * i;
		}
		cout << sum << endl;
	}
}

D

由题意可知，排序规则为区间大小，其次索引大小
递归处理出每个元素的区间大小和索引大小，最后处理出次序即可

pair a[maxn];
int ans[maxn];
void f(int l, int r) {
	if (l == r) {
		a[l] = make_pair(1, l);
		return;
	}
	if (l > r) return;
	if ((r - l + 1) & 1) {
		int idx = (l + r) / 2;
		a[idx] = make_pair(r - l + 1, idx);
		f(l, idx - 1);
		f(idx + 1, r);
	}
	else {
		int idx = (l + r - 1) / 2;
		a[idx] = make_pair(r - l + 1, idx);
		f(idx + 1, r);
		f(l, idx - 1);
	}
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t; cin >> t;
	while (t--) {
		int n; cin >> n;
		f(1, n);
		sort(a + 1, a + n + 1, [&](pair x, pair y) {
			if (x.first != y.first) return x.first > y.first;
			return x.second < y.second;
		});
		for (int i = 1; i <= n; ++i) {
			ans[a[i].second] = i;
		}
		for (int i = 1; i <= n; ++i) {
			cout << ans[i] << ' ';
		}
		cout << endl;
	}
}

E

处理出下标%k的子序列，答案即为选择一个子序列形成000011…1110000这种状态的改变量 与 将其他子序列变为全0 的操作数和的最小值
对每个子序列，考虑字符为’1’是代价为1，字符为’0’代价为-1，即为求最长上升子序列

string str;
vector a;
int main()
{
	ios::sync_with_stdio(false); cin.tie(0);
	int t; cin >> t;
	while (t--) {
		int n, k; cin >> n >> k >> str;
		int sum = 0, ans = 1e9;
		for (int i = 0; i < n; ++i) sum += str[i] == '1';
		for (int i = 0; i < k; ++i) {
			a.clear(); 
			int tmp = 0, now = 0;
			for (int j = i; j < n; j += k) {
				if (str[j] == '1') a.push_back(1);
				else a.push_back(-1);
			}
			for (int p = 0; p < a.size(); ++p) {
				tmp += a[p];
				if (tmp < 0) tmp = 0;
				now = max(now, tmp);
			}
			ans = min(ans, sum - now);
			
		}
		cout << ans << endl;
	}
}