P5221 Product

推式子

$$\begin{gathered} \prod _{i=1}^n\prod _{j=1}^n\frac{lcm(i,j)}{gcd(i,j)} \\ \prod _{i=1}^n\prod _{j=1}^n\frac{ij}{gcd(i,j{)}^2} \\ 我們考慮上面 \\ \prod _{i=1}^n\prod _{j=1}^{n}ij \\ \prod _{i=1}^n{i}^n\prod _{j=1}^{n}j \\ \prod _{i=1}^n{i}^{n}n! \\ n{!}^n\prod _{i=1}^n{i}^n \\ 最后得到n{!}^{2n} \\ 再考慮下面化簡(jiǎn) \\ \prod _{i=1}^n\prod _{j=1}^{n}gcd(i,j{)}^2 \\ \prod _{d=1}^n{d}^{2\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}[gcd(i,j)=1]} \\ 對(duì)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}[gcd(i,j)==1]化簡(jiǎn) \\ \sum _{k=1}^{\frac{n}{d}}\mu (k){\left(\frac{n}{kd}\right)}^2 \\ 整體化簡(jiǎn)后\frac{n{!}^{2n}}{{\prod }_{d=1}^n{d}^{2\sum _{k=1}^{\frac{n}{d}}\mu (k){\left(\frac{n}{kd}\right)}^2}} \end{gathered}$$

代碼

開c++17,不加$O_2$能過(guò)，卡內(nèi)存，，，

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int mod = 104857601, Mod = mod - 1, N = 1e6 + 10;int mu[N], prime[N], cnt, n;bool st[N];ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }ll inv(ll x) {return quick_pow(x, mod - 2); }void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;}for(int j = 0; j < cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {mu[i] += mu[i - 1];} }ll calc1(ll l, ll r) {ll ans = 1;for(int i = l; i <= r; i++) {ans = ans * i % mod;}return ans; }ll calc2(ll n) {ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll *(mu[r] - mu[l - 1]) * (n / l) % Mod * (n / l) % Mod) % Mod;}return (ans % Mod + Mod) % Mod; }int main() {init();scanf("%d", &n);ll a = 1, b = 1;for(int i = 1; i <= n; i++) a = a * i % mod;a = quick_pow(a, (2 * n) % Mod);for(int l = 1, r; l <= n; l = r + 1) {r = n / (n / l);b = b * quick_pow(calc1(l, r), 2ll * calc2(n / l) % Mod) % mod;}printf("%lld\n", 1ll * a * inv(b) % mod);return 0; }

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