Educational Codeforces Round 174 (Rated for Div. 2)

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A - Was there an Array?#

题意#

思路#

有“$101$”就无解

代码#

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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
typedef pair<int, int> pii;

const int mxn = 1e6 + 5;

void solve()
{
	int n;
	cin >> n;
	vector<int> b(n - 2);
	for (int i = 0; i < n - 2; i++)
	{
		cin >> b[i];
	}
	if (n <= 4)
	{
		cout << "YES" << endl;
		return;
	}
	for (int i = 0; i < n - 4; i++)
	{
		if (b[i] == 1 && b[i + 1] == 0 && b[i + 2] == 1)
		{
			cout << "NO" << endl;
			return;
		}
	}
	cout << "YES" << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);

	int __ = 1;
	cin >> __;
	while (__--)
	{
		solve();
	}

	return 0;
}

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B - Set of Strangers#

题意#

给定$n\times m$的网格，每个网格有对应的颜色。每次操作可以选择颜色相同且不相邻的任意网格将其染成一种颜色。问将所有网格染成相同的颜色的最少操作数

思路#

对于某一种颜色，若每个网格都不相邻，则染成其他颜色只需要$1$次操作；有相邻的要$2$次。统计每种颜色是否相邻，无贡献为$1$，否则为$2$，最后选定要染的颜色——为了减少操作数，尽量选择有相邻的颜色

代码#

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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
typedef pair<int, int> pii;

const int mxn = 2e5 + 10;

int dx[] = { 1,-1,0,0 };
int dy[] = { 0,0,1,-1 };

void solve()
{
	int n, m;
	cin >> n >> m;
	unordered_set<int> s;
	vector<bool> connect(n * m + 1, false);
	vector<vector<int>> a(n, vector<int>(m));
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			cin >> a[i][j];
			s.insert(a[i][j]);
		}
	}
	if (s.size() == 1)
	{
		cout << 0 << endl;
		return;
	}
	bool f = false;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < m; j++)
		{
			if (connect[a[i][j]])
			{
				continue;
			}
			for (int k = 0; k < 4; k++)
			{
				int tx = i + dx[k];
				int ty = j + dy[k];
				if (tx < 0 || tx >= n || ty < 0 || ty >= m || a[tx][ty] != a[i][j])
				{
					continue;
				}
				connect[a[i][j]] = f = true;
				break;
			}
		}
	}
	int ans = 0;
	for (auto& i : s)
	{
		ans += (connect[i] ? 2 : 1);
	}
	cout << ans - 1 - f << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);

	int __ = 1;
	cin >> __;
	while (__--)
	{
		solve();
	}

	return 0;
}

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C - Beautiful Sequence#

题意#

思路#

由于只含有$1,2,3$三种元素，所以合法序列只能是$122···223$的形式。所以答案即对于每一对$(1,3)$，求出其中间$2$的个数$m$，这对$(1,3)$的贡献就是$\sum _{i=1}^m{C}_m^i$。再预处理一下，避免超时

代码#

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#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long

const int mod = 998244353;
const int mxn = 2e5 + 10;

int pow2[mxn], inv_pow2[mxn];

int qpow(int a, int b)
{
	int res = 1;
	while (b)
	{
		if (b & 1)
		{
			res = res * a % mod;
		}
		a = a * a % mod;
		b >>= 1;
	}
	return res;
}

void init()
{
	pow2[0] = inv_pow2[0] = 1;
	int inv_2 = qpow(2, mod - 2);
	for (int i = 1; i < mxn; i++)
	{
		pow2[i] = pow2[i - 1] * 2 % mod;
		inv_pow2[i] = inv_pow2[i - 1] * inv_2 % mod;
	}
}



void solve()
{
	int n;
	cin >> n;
	vector<int> a(n), cnt2(n);
	vector<int> p1, p3;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		if (a[i] == 2)
		{
			cnt2[i] = (i ? cnt2[i - 1] + 1 : 1);
		}
		else
		{
			cnt2[i] = (i ? cnt2[i - 1] : 0);
			if (a[i] == 1)
			{
				p1.push_back(i);
			}
			else if (a[i] == 3)
			{
				p3.push_back(i);
			}
		}
	}
	if (p1.empty() || p3.empty())
	{
		cout << 0 << endl;
		return;
	}
	vector<int> pre(1, 0);
	for (int i : p1)
	{
		int ai = cnt2[i];
		pre.push_back((pre.back() + inv_pow2[ai]) % mod);
	}
	int ans = 0;
	for (int i : p3) 
	{
		int cnt = cnt2[i];
		int num = upper_bound(p1.begin(), p1.end(), i) - p1.begin();
		if (num == 0) 
		{
			continue;
		}
		int sum = pre[num];
		int t = (pow2[cnt] * sum % mod - num + mod) % mod;
		ans = (ans + t) % mod;
	}
	cout << ans << endl;
}

signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);

	int __ = 1;
	cin >> __;
	init();
	while (__--)
	{
		solve();
	}

	return 0;
}

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D - Palindrome Shuffle#

题意#

思路#

代码#

点击查看代码

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比赛链接 https://mirror.codeforces.com/contest/2069