洛谷 P10084 [GDKOI2024 提高组] 计算

洛谷传送门

第一步是一个经典结论，\(L = m^{\gcd(a, b)} + 1\)，\(R = m^{\gcd(c, d)}\)。

因为 \(L \equiv 1 \pmod m\) 且 \(R \equiv 0 \pmod m\)，所以可以把问题的范围改成 \([1, n = R - L + 1]\)。

写出选数的生成函数：

\[F(x) = \prod\limits_{i = 1}^n (1 + x^i) \]

我们希望求所有次数是 \(m\) 的倍数的项的系数之和。

施单位根反演：

\[\begin{aligned} ans & = \sum\limits_{i \ge 0} [m \mid i] [x^i] F(x) \\ & = \frac{1}{m} \sum\limits_{i \ge 0} \sum\limits_{j = 0}^{m - 1} \omega_m^{ij} [x^i] F(x) \\ & = \frac{1}{m} \sum\limits_{j = 0}^{m - 1} \sum\limits_{i \ge 0} (\omega_m^j)^i [x^i] F(x) \\ & = \frac{1}{m} \sum\limits_{j = 0}^{m - 1} F(\omega_m^j) \end{aligned} \]

考虑对于一个给定的 \(j\) 如何计算 \(F(\omega_m^j)\) 即 \(\prod\limits_{i = 1}^n (1 + (\omega_m^j)^i)\)。

设 \(d = \gcd(m, j)\)。因为有 \(\omega_m^k = \omega_m^{k \bmod m}\)，又因为 \(\frac{j}{d}, \frac{2j}{d}, \ldots, \frac{nj}{d}\) 形成了 \(\frac{n}{\frac{m}{d}}\) 个模 \(\frac{m}{d}\) 的剩余系，所以：

\[\begin{aligned} F(\omega_m^j) & = \prod\limits_{i = 1}^n (1 + (\omega_{\frac{m}{d}}^{\frac{j}{d}})^i) \\ & = \prod\limits_{k = 1}^{\frac{m}{d}} (1 + \omega_{\frac{m}{d}}^k)^{\frac{n}{\frac{m}{d}}} \\ & = \prod\limits_{k = 0}^{\frac{m}{d} - 1} (1 + \omega_{\frac{m}{d}}^k)^{\frac{n}{\frac{m}{d}}} \end{aligned} \]

考虑求这样一个式子：\(\prod\limits_{i = 0}^{n - 1} (1 + \omega_n^i)\)。考虑分圆多项式：

\[x^n - 1 = \prod\limits_{i = 0}^{n - 1} (x - \omega_n^i) \]

代入 \(x = -1\) 得：

\[(-1)^n - 1 = \prod\limits_{i = 0}^{n - 1} (-1 - \omega_n^i) \]

\[1 - (-1)^n = \prod\limits_{i = 0}^{n - 1} (1 + \omega_n^i) \]

也就是当 \(\frac{m}{d}\) 为偶数时，\(F(\omega_m^j) = 0\)；否则 \(F(\omega_m^j) = 2^{\frac{n}{\frac{m}{d}}}\)。

把 \(F(\omega_m^j)\) 代入到答案的式子中，有：

\[ans = \frac{1}{m} \sum\limits_{j = 0}^{m - 1} [2 \nmid \frac{m}{\gcd(m, j)}] 2^{\frac{n}{\frac{m}{\gcd(j, m)}}} \]

直接计算，时间复杂度 \(O(Tm \log n)\)，无法通过。

考虑一些 trivial 的优化，枚举 \(d = \gcd(m, j)\)，\([0, m - 1]\) 中和 \(m\) 的 \(\gcd\) 为 \(d\) 的数的个数显然为 \(\varphi(\frac{m}{d})\)，那么：

\[ans = \frac{1}{m} \sum\limits_{d \mid m} [2 \nmid \frac{m}{d}] 2^{\frac{n}{\frac{m}{d}}} \varphi(\frac{m}{d}) \]

先线性筛出全部 \(\varphi(i)\) 即可计算。时间复杂度降为 \(O(T(\sqrt{m} + d(m) \log n))\)，可以通过。

code

// Problem: P10084 [GDKOI2024 提高组] 计算
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P10084
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 10000100;
const int N = 10000000;
const ll mod = 998244353;

ll n, m, pr[maxn / 10], tot, phi[maxn];
bool vis[maxn];

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

inline void init() {
	phi[1] = 1;
	for (int i = 2; i <= N; ++i) {
		if (!vis[i]) {
			pr[++tot] = i;
			phi[i] = i - 1;
		}
		for (int j = 1; j <= tot && i * pr[j] <= N; ++j) {
			vis[i * pr[j]] = 1;
			if (i % pr[j] == 0) {
				phi[i * pr[j]] = phi[i] * pr[j];
				break;
			}
			phi[i * pr[j]] = phi[i] * (pr[j] - 1);
		}
	}
}

inline ll work(ll d) {
	if ((m / d) % 2 == 0) {
		return 0;
	}
	return qpow(2, n / (m / d)) * phi[m / d] % mod;
}

void solve() {
	ll _a, _b, _c, _d;
	scanf("%lld%lld%lld%lld%lld", &m, &_a, &_b, &_c, &_d);
	ll _x = 1, _y = 1;
	for (int _ = 0; _ < __gcd(_c, _d); ++_) {
		_x *= m;
	}
	for (int _ = 0; _ < __gcd(_a, _b); ++_) {
		_y *= m;
	}
	n = _x - _y;
	if (!_a || !_b) {
		n = _x;
	}
	ll ans = 0;
	for (ll i = 1; i * i <= m; ++i) {
		if (m % i) {
			continue;
		}
		ans = (ans + work(i)) % mod;
		if (i * i != m) {
			ans = (ans + work(m / i)) % mod;
		}
	}
	printf("%lld\n", ans * qpow(m, mod - 2) % mod);
}

int main() {
	init();
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}