2024.2.7-8 寒假训练记录（21）

洛谷P3193 [HNOI2008] GT考试

题目链接

KMP+dp+矩阵快速幂

还没有理解得很清楚，主要是对KMP理解还不够深刻

#include 

using namespace std;

#define int long long
using i64 = long long;

typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;
typedef pair<PII, PII> PIIII;

const int N = 2010;

int n, m, mod;

struct Martix
{
	int a[30][30]; // 在这里修改矩阵的大小
	Martix() { memset(a, 0, sizeof(a)); }
	Martix operator*(const Martix &B) const // 乘法运算符重载
	{
		Martix res;
		for (int i = 0; i < m; i ++ )
			for (int j = 0; j < m; j ++ )
				for (int k = 0; k < m; k ++ )
					res.a[i][j] = (res.a[i][j] + a[i][k] * B.a[k][j]) % mod;
		return res;
	}
} G;

vector<int> pi(N);
void get_pi(string s)
{
	int j = 0;
	for (int i = 2; i <= m; i++)
	{
		while (j && s[j + 1] != s[i]) j = pi[j];
		if (s[j + 1] == s[i]) j ++ ;
		pi[i] = j;
	}
	for (int i = 0; i < m; i++)
	{
		for (char ch = '0'; ch <= '9'; ch++)
		{
			j = i;
			while (j && s[j + 1] != ch) j = pi[j];
			if (s[j + 1] == ch) j ++ ;
			G.a[i][j] ++ ;
		}
	}
}

Martix power(Martix &a, int b)
{
	Martix ans;
	for (int i = 0; i < m; i ++ ) ans.a[i][i] = 1;
	while (b)
	{
		if (b & 1) ans = ans * a;
		b >>= 1;
		a = a * a;
	}
	return ans;
}

void solve()
{
	cin >> n >> m >> mod;
	string s; cin >> s;
	s = " " + s;
	get_pi(s);
	G = power(G, n);
	int ans = 0;
	for (int i = 1; i <= m; i ++ ) ans = (ans + G.a[0][i]) % mod;
	cout << ans << '\n';
}

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

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

ATC abc339E Smooth Subsequence

题目链接

线段树优化dp

把以每个数字结尾的最佳答案存进线段树中，查询即可

#include 

using namespace std;

#define int long long
using i64 = long long;

typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;

const int N = 5e5 + 10;
const int mod = 1e9 + 7;

struct Node
{
	int l, r, maxx;
} tr[N * 4];

void pushup(Node &u, Node &left, Node &right)
{
	u.maxx = max(left.maxx, right.maxx);
}

void pushup(int u)
{
	pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

void build(int u, int l, int r)
{
	tr[u] = {l, r, 0};
	if (l == r) return;
	int mid = l + r >> 1;
	build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
}

void modify(int u, int pos, int x)
{
	if (tr[u].l == pos && tr[u].r == pos)
	{
		tr[u].maxx = x;
		return;
	}
	int mid = tr[u].l + tr[u].r >> 1;
	if (pos <= mid) modify(u << 1, pos, x);
	else modify(u << 1 | 1, pos, x);
	pushup(u);
}

Node query(int u, int l, int r)
{
	if (tr[u].l >= l && tr[u].r <= r) return tr[u];
	int mid = tr[u].l + tr[u].r >> 1;
	if (r <= mid) return query(u << 1, l, r);
	else if (l > mid) return query(u << 1 | 1, l, r);
	else
	{
		auto left = query(u << 1, l, mid);
		auto right = query(u << 1 | 1, mid + 1, r);
		Node res = {l, r};
		pushup(res, left, right);
		return res;
	}
}

void solve()
{
	int n, d;
	cin >> n >> d;
	build(1, 1, N);
	vector<int> dp(n + 1);
	for (int i = 1; i <= n; i ++ )
	{
		int x;
		cin >> x;
		Node res = query(1, x - d, x + d);
		dp[i] = max((i64)1, res.maxx + 1);
		modify(1, x, dp[i]);
	}
	int ans = 0;
	for (int i = 1; i <= n; i ++ )
	{
		ans = max(ans, dp[i]);
	}
	cout << ans << '\n';
}

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

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

ATC abc339F Product Equality

题目链接

哈希

这里的一个trick是，乘积的个数比较少，哈希之后很可能出现一样的关键字，此时可以进行双哈希（或更多的哈希都没问题），取不同的模数，减小出现相同关键字的概率

#include 

using namespace std;

#define int long long
using i64 = long long;

typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<int, PII> PIII;

const int N = 5e5 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;

void solve()
{
	int n;
	cin >> n;
	vector<string> a(n);
	map<int, int> mp1, mp2;
	auto mm = [&](string s, int mod)
	{
		int res = 0;
		for (auto t : s)
		{
			res = (res * 10 + t - '0') % mod;
		}
		return res;
	};
	vector<int> b1(n), b2(n);
	for (int i = 0; i < n; i ++ )
	{
		cin >> a[i];
		b1[i] = mm(a[i], mod1);
		mp1[b1[i]] ++ ;
		b2[i] = mm(a[i], mod2);
		mp2[b2[i]] ++ ;
	}
	int ans = 0;
	for (int i = 0; i < n; i ++ )
	{
		for (int j = 0; j < n; j ++ )
		{
			ans += min(mp1[(b1[i] * b1[j]) % mod1], mp2[(b2[i] * b2[j]) % mod2]);
		}
	}
	cout << ans << '\n';
}

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

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