P4917 天守阁的地板

P4917 天守阁的地板

写得我头疼。

容易发现，任意两个数要拼成一个正方形，最小边长是 (operatorname{lcm}(i,j))

那么需要用的数量就是 (dfrac{operatorname{lcm}(i,j)^2}{i cdot j})

所以我们要求的就是 (displaystyle prod_{i=1}^N prod_{j=1}^N dfrac{operatorname{lcm}(i, j)^2}{i cdot j})

转化一下就是 (displaystyle prod_{i=1}^N prod_{j=1}^N dfrac{i cdot j}{gcd(i,j)^2})

接下来就是 (displaystyle dfrac{large (n!)^{2n}}{displaystyle prod_{i=1}^N prod_{j=1}^N gcd(i, j)^2})

然后我们令 (d = gcd(i, j))

那么 (displaystyle dfrac{large (n!)^{2n}}{displaystyle prod_{d=1}^N d^{2 cdot 2 sum_{i=1}^{lfloor frac{N}{d} floor } sum_{j=1}^{lfloor frac{N}{d} floor }[gcd(i,j)=1]}})

然后经过莫比乌斯反演（懒得写了）

然后得到 (displaystyle dfrac{large (n!)^{2n}}{displaystyle prod_{d=1}^N d^{2 cdot 2 sum_{k=1}^{lfloor frac{N}{d} floor} mu(k) cdot (lfloor frac{N}{kd} floor )^2}})

感觉是可以做了，但是我写挂了，然后换了个方法。。。

你考虑这就是求了一个数表内互质的数

那么可以换成

[huge displaystyle dfrac{ (n!)^{2n}}{displaystyle prod_{d=1}^N d^{2 cdot Big(2 cdot sum_{i=1}^{lfloor frac{N}{d} floor} varphiig(iig)Big) - 1}} ]

注：(-1) 是因为 (varphi(1)) 被算了两遍。

下面就是各位的代码能力了。。。

---

下面是我的代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>

using namespace std;

typedef long long ll;
const ll MAXN = 1e6+10, MOD = 19260817;

ll N, T, fac[MAXN], phi[MAXN], prime[MAXN], cnt, vis[MAXN], inv[MAXN];

ll ks(ll, ll);
ll f(ll);

int main() {
    scanf("%lld", &T);
    fac[0] = fac[1] = inv[0] = inv[1] = 1;
    for (ll i = 2; i <= MAXN - 10; i++)
        fac[i] = fac[i-1] * i % MOD, inv[i] = inv[MOD % i] * (MOD - MOD / i) % MOD;
    for (ll i = 2; i <= MAXN - 10; i++)	
    	inv[i] = inv[i-1] * inv[i] % MOD;
    phi[1] = 1;
    for (ll i = 2; i <= MAXN - 10; i++) {
        if (!vis[i]) prime[++cnt] = i, phi[i] = i-1;
        for (ll j = 1; j <= cnt && prime[j] * i <= MAXN - 10; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j]) {
                phi[i * prime[j]] = phi[i] * phi[prime[j]];
            } else {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
        }
    }
    for (ll i = 1; i <= MAXN - 10; i++)
    	phi[i] = (phi[i] + phi[i-1]) % (MOD - 1); // phi出现在指数，那么 mod (MOD-1)
    while (T--) {
        scanf("%lld", &N);
        ll up = ks(fac[N], (2 * N) % MOD) % MOD;
        ll down = 1;
        for (ll l = 1, r = 0; l <= N; l = r + 1) {
            r = (N / (N / l));
            ll gx = phi[N / l] * 2 - 1;
            down = down * ks(fac[r] * inv[l-1] % MOD, gx) % MOD;
        }
        down = down * down % MOD;
        printf("%lld
", up * ks(down, MOD-2) % MOD);
    }
    return 0;
}

ll ks(ll n, ll tim) {
    ll ret = 1;
    while (tim) {
        if (tim & 1) ret = ret * n % MOD;
        n = n * n % MOD;
        tim >>= 1;
    }
    return ret % MOD;
}