Codeforces Round 1006 (Div. 3) 补题+题解

A. New World, New Me, New Array

贪心的想每次都赋值一个 \(p\) 如果正好和为 \(k\) 则答案就是 \(k/p\) ，否则是 \(k/p+1\)。

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;

void solve() {
	int n, k, p;
	cin >> n >> k >> p;
	if (n * p < abs(k)) {
		cout << -1 << '\n';
		return;
	}
	int ans = abs(k / p);
	if (k % p != 0) ans++;
	cout << ans << '\n';
}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--)
		solve();
}

B. Having Been a Treasurer in the Past, I Help Goblins Deceive

简单地想，\(O(n)\) 扫描一遍记录前后各放多少个 "-" 会使子序列最多，直接计算即可。

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;

void solve() {
	string s;
	int n;
	cin >> n;
	cin >> s;
	int cnt = 0, cnt1 = 0;
	for (auto t : s) {
		if (t == '-') cnt++;
	}
	int mx = 0;
	cnt1 = s.size() - cnt;
	for (int i = 1; i <= cnt; i++) {
		int x = i, y = cnt - i;
		mx = max(mx, x * y);
	}
	cout << mx*cnt1 << '\n';
}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--)
		solve();
}

C. Creating Keys for StORages Has Become My Main Skill

贪心 ,\([0,n-1]\) 尽可能取，维护或值，如果恰好为 \(x\) 则是答案。否则尽可能修改大的将最后一项取为 \(x\),这样 \(MEX\) 最大。

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;

void solve() {
	int n, x;
	cin >> n >> x;
	int now = 0;
	vector<int> ans(n+10, 0);
	for (int i = 1; i < n; i ++ ) {
		if ((x | i) == x) {
			ans[i] = i;
			now |= i;
		}
	}
	if (now != x) {
		ans[n - 1] = x;
	}
	for (int i = 0; i < n; i ++ ) {
		cout << ans[i] << ' ';
	}
	cout << '\n';
}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--)
		solve();
}

D. For Wizards, the Exam Is Easy, but I Couldn't Handle It

很明显反转数的个数变化只与选择的区间有关，与区间外的数无关，区间的贡献则是区间中小于 \(l\) 的个数减去大于 \(l\) 的个数 。

\(n^2\) 暴力枚举，在过程中记下 \([l,r]\) 区间中大于 \(l\) 和小于 \(l\) 的个数,然后统计贡献最大的区间便是答案。

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;

void solve() {
	int n;
	cin >> n;
	vector<int> a(n + 1);
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	int ans = 0;
	pair<int, int> anss;
	for (int l = 1; l <= n; l++) {
		int x = 0, y = 0;
		for (int r = l; r <= n; r++) {
			if (a[r] > a[l]) x++;
			if (a[r] < a[l]) y++;
			if (y - x >= ans) {
				ans = y - x;
				anss.first = l, anss.second = r;
			}
		}
	}
	cout << anss.first << ' ' << anss.second << '\n';
}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--)
		solve();
}

E. Do You Love Your Hero and His Two-Hit Multi-Target Attacks?

大致题意：Akito需要在坐标系中放置\(n\) 根法杖，且这些法杖的位置必须是不同的整数坐标点。放置后，恰好有\(k\) 对法杖位于同一条水平线或垂直线上。

构造题，利用组合数的知识 \(C(x,2)\) 尽可能构造即可。

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;

void solve() {
	int n;
	cin >> n;
	if (n == 0) {
		cout << "2\n0 0\n1 1\n";
		return;
	}
	vector<pair<int, int>> ans;
	int yu = n;
	int ls = 1, d = 0;
	while (yu > 2) {
		for (int i = 2; i <= 500; i++) {
			int x = (i - 1) * i / 2;
			if (x > yu) {
				int res = (i - 1) * (i - 2);
				yu -= res / 2;
				for (int j = ls; j <= ls + i - 1 - 1; j++) {
					ans.push_back({-j, d});
				}
				ls = ls + i - 1;
				d -= 1;
				break;
			}
		}
	}
	if (yu == 2) {
		ans.push_back({1, 1});
		ans.push_back({2, 1});
		ans.push_back({3, 2});
		ans.push_back({4, 2});
	} else if (yu == 1) {
		ans.push_back({1, 1});
		ans.push_back({2, 1});
	}
	cout << ans.size() << '\n';
	for (pair<int, int> p : ans) {
		cout << p.first << ' ' << p.second << '\n';
	}
}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--)
		solve();
}

F. Goodbye, Banker Life

打表找一下规律可以发现第 \(n\) 行是由最近的第 \(n-2^x\) 行转移过来的，那么很明显可以递归，通过递归记录一下答案，处理一下直接输出。代码实现比较抽象，具体看代码。

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 10;
int n, x;
int ksm(int x, int y) {
	int ans = 1;
	while (y) {
		if (y & 1)
			ans = ans * x;
		y >>= 1;
		x = x * x;
	}
	return ans;
}
vector<int> po;
void init() {
	for (int i = 32; i >= 0; i--) {
		int x = ksm(2, i);
		po.push_back(x);
	}
}
deque<pair<string, int>> de;
void dfs(int u) {
	if (u == 1 || (u & (u - 1)) == 0) {
		if (u == 1) {
			de.push_front({"1", 1});
		} else {
			de.push_back({"1", u});
		}
		return;
	}
	for (int i = 0; i < 32; i++) {
		if (po[i] <= u) {
			dfs(u - po[i]);
			int yu = u - 2 * (u - po[i]);
			de.push_back({"0", yu});
			de.push_back({"x", 0});
			return;
		}
	}
}
void pr();
void pr1(int l, int r);
void pr0(int l, int r);
void solve() {
	while (de.size()) de.pop_front();
	cin >> n >> x;
	string s = to_string(1);
	for (int i = 0; i < 32; i++) {
		if (n == po[i]) {
			pr();
			return;
		}
	}
	dfs(n);
	vector<pair<string, int>> p;
	while (de.size()) {
		pair<string, int> it = de.front();
		de.pop_front();
		p.push_back(it);
	}
	for (int i = 0; i < p.size(); i++) {
		string a = p[i].first;
		int b = p[i].second;
		if (a == "0") {
			pr0(1, b);
		}
		else if (a == "1") {
			pr1(1, b);
		}
		else if (a == "x") {
			int len = 0;
			if (p[i - 1].first == "0") len = i - 1;
			else len = i;
			string lp = "";
			for (int j = 0; j < len; j++) {
				int cnt = p[j].second;
				if (p[j].first == "0")
					for (int k = 1; k <= cnt; k++) {
						cout << 0 << ' ';
						lp += "0";
					}
				else if (p[j].first == "1") {
					for (int k = 1; k <= cnt; k++) {
						cout << x << ' ';
						lp += "1";
					}
				}
				else
					for (char t : p[j].first) {
						if (t == '1') cout << x << ' ', lp += "1";
						else if (t == '0') cout << 0 << ' ', lp += "0";
					}
			}
			p[i].first = lp;
		}
	}
	cout << '\n';
}
void pr() {
	for (int i = 1; i <= n; i++)
		cout << x << ' ';
	cout << '\n';
}
void pr1(int l, int r) {
	for (int i = l; i <= r; i++) cout << x << ' ';
}
void pr0(int l, int r) {
	for (int i = l; i <= r; i++) cout << 0 << ' ';
}

signed main() {
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int t = 1;
	init();
	cin >> t;
	while (t--)
		solve();
}
/*
第 1 行: 1
第 2 行: 1 1
第 3 行: 1 0 1
第 4 行: 1 1 1 1
第 5 行: 1 0 0 0 1
第 6 行: 1 1 0 0 1 1
第 7 行: 1 0 1 0 1 0 1
第 8 行: 1 1 1 1 1 1 1 1
第 9 行: 1 0 0 0 0 0 0 0 1
第 10行: 1 1 0 0 0 0 0 0 1 1
*/