题目地址

题目链接

题解

注,下方\((i,j)\)均指\(gcd(i,j)\),以及证明过程有一定的跳步,请确保自己会莫比乌斯反演的基本套路。

介绍本题的\(O(n)\)和\(O(n\sqrt{n})\)做法,本题还有\(O(nlogn)\)做法,需要用到欧拉函数,或者是从质因子角度考虑也可以得到另外一个\(O(n)\)做法。

题目就是求

\[\prod_{i=1}^n\prod_{j=1}^n\frac{ij}{(i,j)^2}
\]

考虑分解一下

\[\prod_{i=1}^n\prod_{j=1}^n\frac{ij}{(i,j)^2}=\frac{\prod_{i=1}^n\prod_{j=1}^nij}{\prod_{i=1}^n\prod_{j=1}^n(i,j)^2}
\]

对于分子可得

\[\begin{aligned}
&\prod_{i=1}^n\prod_{j=1}^nij\\
&=\prod_{i=1}^ni\prod_{j=1}^nj\\
&=\prod_{i=1}^ni*n!\\
&=(n!)^{2n}
\end{aligned}
\]

对于分母,我们考虑莫比乌斯反演

\[\begin{aligned}
&\prod_{i=1}^n\prod_{j=1}^n(i,j)^2\\
&=\prod_{d=1}^nd^{2\sum_{i=1}^n\sum_{j=1}^n[(i,j)=d]}\\
&=\prod_{d=1}^nd^{2\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}[(i,j)=1]}\\
&=\prod_{d=1}^nd^{2\sum_{k=1}^{\lfloor\frac{n}{d}\rfloor}\mu(k)\lfloor\frac{n}{kd}\rfloor^2}\\
\end{aligned}
\]

至此,枚举\(d\),对指数整除分块,即可\(O(n\sqrt{n})\)解决此题。

容易发现\(\lfloor\frac{n}{d}\rfloor\)是可以整除分块的。那么怎么处理区间\([l,r]\)的\(d\)呢,将它展开,其实就是\(\frac{r!}{(l-1)!}\),由于出题人卡空间,所以可以直接计算阶乘而不是预处理(复杂度同样是\(O(n)\),每个数只会被遍历一次)

那么就可以做到\(O(n)\)解决本题了。

#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std; const int mod = 104857601;
const int p = 104857600;
const int N = 1000010; bool vis[N];
short mu[N];
int pr[N], cnt = 0;
int fac; int power(int a, int b, int Mod) {
int ans = 1;
while(b) {
if(b & 1) ans = (ll)ans * a % Mod;
a = (ll)a * a % Mod;
b >>= 1;
}
return ans % Mod;
} void init(int n) {
mu[1] = 1;
for(int i = 2; i <= n; ++i) {
if(!vis[i]) pr[++cnt] = i, mu[i] = -1;
for(int j = 1; j <= cnt && i * pr[j] <= n; ++j) {
vis[i * pr[j]] = 1;
if(i % pr[j] == 0) break;
mu[i * pr[j]] = -mu[i];
}
mu[i] += mu[i - 1];
}
fac = 1;
for(int i = 1; i <= n; ++i) fac = (ll)fac * i % mod;
} int n; int calc2(int n) {
int ans = 0;
for(int l = 1, r; l <= n; l = r + 1) {
r = n / (n / l);
ans = (ans + (ll)(n / l) * (n / l) % p * (mu[r] - mu[l - 1] + p) % p) % p;
}
return ans % p;
} int main() {
scanf("%d", &n);
init(n);
int ans = 1;
int sum = power((ll)fac * fac % mod, n, mod);
for(int l = 1, r; l <= n; l = r + 1) {
r = n / (n / l); fac = 1ll;
for(int i = l; i <= r; ++i) fac = (ll)fac * i % mod;
int t = power((ll)fac * fac % mod, calc2(n / l), mod);
ans = (ll)ans * t % mod;
}
printf("%lld\n", (ll)sum * power(ans, mod - 2, mod) % mod);
}

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