BZOJ4833: [Lydsy1704月赛]最小公倍佩尔数
Problem
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;
}
