BZOJ 4816. [Sdoi2017]数字表格

 

$$\prod_{i=1}^{n}\prod_{j=1}^{m}f(\gcd(i,j))$$
$$=\prod_{d=1}^{\min\{n,m\}}f(d)^{\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[(i,j)=1]}$$
$$=\prod_{d=1}^{\min\{n,m\}}f(d)^{\sum_{d'}\mu(d')\lfloor\frac{n}{dd'}\rfloor\lfloor\frac{m}{dd'}\rfloor}$$
$$=\prod_{T}(\prod_{d|T}f(d)^{\mu(\frac{T}{d})})^{\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor}$$
求出 $g(n) =\prod_{d|n}f(d)^{\mu(\frac{n}{d})}$ 的前缀积即可。

#include <bits/stdc++.h>

const int MOD = 1e9 + 7;

int qp(int a, int b = MOD - 2) {
    int ans = 1;
    while (b) {
        if (b & 1) ans = 1LL * ans * a % MOD;
        b >>= 1;
        a = 1LL * a * a % MOD;
    }
    return ans;
}

const int N = 1e6;
int prime[N + 7], prin, mu[N + 7], g[N + 7], f[N + 7], fi[N + 7];
bool vis[N + 7];

inline int add(int x, int y) {
    return x + y >= MOD ? x + y - MOD : x + y;
}

void init() {
    mu[1] = 1;
    f[0] = 0; f[1] = 1;
    fi[1] = 1;
    g[0] = g[1] = 1;
    for (int i = 2; i <= N; i++) {
        g[i] = 1;
        f[i] = add(f[i - 1], f[i - 2]);
        fi[i] = qp(f[i]);
        if (!vis[i]) {
            prime[++prin] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= prin && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= N; i++) if (mu[i])
        for (int j = 1; 1LL * i * j <= N; j++)
            g[i * j] = 1LL * g[i * j] * (mu[i] == 1 ? f[j] : fi[j]) % MOD;
    for (int i = 2; i <= N; i++)
        g[i] = 1LL * g[i] * g[i - 1] % MOD;
}

int main() {
    init();
    int T;
    scanf("%d", &T);
    while (T--) {
        int n, m;
        scanf("%d%d", &n, &m);
        if (n > m) std::swap(n, m);
        int ans = 1;
        for (int i = 1, j; i <= n; i = j + 1) {
            j = std::min(n / (n / i), m / (m / i));
            ans = 1LL * ans * qp(1LL * g[j] * qp(g[i - 1]) % MOD, 1LL * (n / i) * (m / i) % (MOD - 1)) % MOD;
        }
        printf("%d\n", ans);
    }
    return 0;
}