BZOJ4833: [Lydsy1704月赛]最小公倍佩尔数

Problem

传送门

Sol

容易得到

\[f_n=e_{n-1}+f_{n-1},e_{n-1}=f_{n-1}+e_{n-1},f_1=e_1=1
\]

那么

\[f_n=2\times \sum_{i=1}^{n-1}f_i-f_{n-1}+1
\]

又有

\[f_{n+1}=2\times \sum_{i=1}^{n}f_i-f_{n}+1
\]

相减得到 \(f_{n+1}=f_n\times 2 + f_{n-1},f_1=1\)

有结论 \(gcd(a,b)=1\) 时，形如 \(f_i=af_{i-1}+bf_{i-2}\) 的数列有性质 \(gcd(f_i,f_j)=f_{gcd(i,j)}\)

大概可以这么证明

而 \(lcm\) 实际上是一个对于质因子的指数取 \(max\) 的操作

每个质因子分开考虑，然后最值反演

\[lcm(S)=\prod_{i}p_i^{\sum_{T\subset S}min(T_i)(-1)^{|T|+1}}=\prod_{i}\prod_{T\subset S}p_i^{min(T_i)(-1)^{|T|+1}}
\]

交换顺序得到

\[lcm(S)=\prod_{T \subset S}gcd(T)^{(-1)^{|T|+1}}
\]

其中 \(S,T\ne \phi\)，\(min(T_i)\) 表示 \(p_i\) 这个因子的指数最小值

所以

\[g_n=\prod_{T \subset S}f_{gcd(T)}^{(-1)^{|T|+1}}=\prod_{d=1}^{n}f_{d}^{\sum_{T\subset S}[gcd(T)==d](-1)^{|T|+1}}
\]

设 $$s_d=\sum_{T\subset S}gcd(T)==d^{|T|+1}$$

\[h_i=\sum_{i|d}^{n}s_d=\sum_{T\subset S}[i|gcd(T)](-1)^{|T|+1}=[cnt_i\ne 0]=1
\]

其中 \(cnt_i\) 表示 \(i\) 的倍数的个数

那么

\[s_i=\sum_{i|d}^{n}\mu(\frac{d}{i})h_d=\sum_{i|d}^{n}\mu(\frac{d}{i})
\]

那么

\[g_n=\prod_{d=1}^{n}f_{d}^{\sum_{d|i}^{n}\mu(\frac{i}{d})}=\prod_{d=1}^{n}\prod_{d|i}^{n}f_d^{\mu(\frac{i}{d})}=\prod_{i=1}^{n}\prod_{d|i}f_d^{\mu(\frac{i}{d})}
\]

\(\Theta(nlog n)\) 预处理出 \(\prod_{d|i}f_d^{\mu(\frac{i}{d})}\) 即可

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(1e6 + 5);

int pr[maxn], num, ispr[maxn], mu[maxn], n, f[maxn], s[maxn], g[maxn], mod, ans, inv[maxn];

inline int Pow(ll x, int y) {
	register ll ret = 1;
	for (; y; y >>= 1, x = x * x % mod)
		if (y & 1) ret = ret * x % mod;
	return ret;
}

inline void Inc(int &x, int y) {
	if ((x += y) >= mod) x -= mod;
}

int main() {
	register int i, j, test;
	mu[1] = 1, ispr[1] = 1;
	for (i = 2; i <= 1000000; ++i) {
		if (!ispr[i]) pr[++num] = i, mu[i] = -1;
		for (j = 1; j <= num && i * pr[j] <= 1000000; ++j) {
			ispr[i * pr[j]] = 1;
			if (i % pr[j]) mu[i * pr[j]] = -mu[i];
			else {
				mu[i * pr[j]] = 0;
				break;
			}
		}
	}
	mod = 1e9 + 7;
	for (scanf("%d", &test); test; --test) {
		scanf("%d%d", &n, &mod), ans = 0;
		for (f[1] = 1, i = 2; i <= n; ++i) f[i] = (2LL * f[i - 1] + f[i - 2]) % mod;
		for (g[0] = i = 1; i <= n; ++i) g[i] = 0, s[i] = 1, inv[i] = Pow(f[i], mod - 2);
		for (i = 1; i <= n; ++i)
			for (j = i; j <= n; j += i)
				if (mu[j / i] == 1) s[j] = 1LL * s[j] * f[i] % mod;
				else if (mu[j / i] == -1) s[j] = 1LL * s[j] * inv[i] % mod;
		for (i = 1; i <= n; ++i) g[i] = 1LL * g[i - 1] * s[i] % mod;
		for (i = 1; i <= n; ++i) Inc(ans, 1LL * i * g[i] % mod);
		printf("%d\n", ans);
	}
    return 0;
}