[SDOI2017]数字表格

$m\\ \prod_{i = 1} ^{n} \prod_{j = 1} ^{m} f(gcd(i, j))\\ \prod_{d = 1} ^{n} f(d) ^{\sum\limits_{i = 1} ^{n} \sum\limits_{j = 1} ^{m} [gcd(i, j) = d]}\\ \prod_{d = 1} ^{n} f(d) ^{\sum\limits_{i = 1} ^{\frac{n}{d}} \sum\limits_{j = 1} ^{\frac{m}{d}}[gcd(i, j) = 1]}\\ \prod_{d = 1} ^{n} f(d) ^{\sum\limits_{k = 1} ^{\frac{n}{d}} \mu(k) \frac{n}{kd} \frac{m}{kd}}\\ 由于\frac{n}{d}具有分块性质，并且n，m不大，所以我们可以先预处理出斐波那契数列的前缀积，然后进行数论分块即可。\\ 注意因为\sum\limits_{k = 1} ^{\frac{n}{d}} \mu(k) \frac{n}{kd} \frac{m}{kd}是指数形式，所以我们要对其模上(mod - 1)$

交了一发发现 $T L E$ 了， $O (n)$ 的复杂度确实过不了极限数据，，，继续推式子。
$id\\ \prod_{T = 1} ^{n} \prod_{d \mid T} f(d)^{\frac{n}{T} \frac{m}{T} \mu(\frac{T}{d})}\\ 舍去\frac{n}{T} \frac{m}{T}的干扰\\ \prod_{T = 1} ^{n}\left(\prod_{d \mid T} f(d) ^{\mu(\frac{T}{d})} \right) ^{\frac{n}{T} \frac{m}{T}}\\ 所以我们只要预先处理出\prod_{d \mid T} f(d) ^{\mu(\frac{T}{d})}，就能\sqrt n 进行求解了。$

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 1e9 + 7, Mod = mod - 1;ll prime[N], mu[N], fib[N], phi[N], inv_fib[N], f[N], cnt;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 a) {return quick_pow(a, mod - 2);
}void init() {fib[1] = mu[1] = f[1] = 1;for(int i = 2; i < N; i++) {f[i] = 1;fib[i] = (fib[i - 1] + fib[i - 2]) % mod;if(!st[i]) {prime[++cnt] = i;mu[i] = -1;}for(int j = 1; 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++) {inv_fib[i] = inv(fib[i]);}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {if(mu[j / i] == 1) {f[j] = f[j] * fib[i] % mod;}else if(mu[j / i] == -1) {f[j] = f[j] * inv_fib[i] % mod;}}}f[0] = 1;for(int i = 1; i < N; i++) {f[i] = f[i] * f[i - 1] % mod;}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%d", &T);while(T--) {int n, m;scanf("%d %d", &n, &m);if(n > m) swap(n, m);ll ans = 1;for(int l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));ans = ans * quick_pow(f[r] * inv(f[l - 1]) % mod, 1ll * (n / l) * (m / l) % Mod) % mod;}printf("%lld\n", ans);}return 0;
}