题解P3704 [SDOI2017]数字表格

P3704 [SDOI2017]数字表格

直接开始推式子,令 \(k=min(n,m)\)

\[ans=\prod_{i=1}^n\prod_{j=1}^mf_{gcd(i,j)}=\prod_{d=1}^kf_d^{\sum\limits_{i=1}^n\sum\limits_{j=1}^m[gcd(i,j)=d]} \]

上面那坨非常明显的莫反

\[=\prod_{d=1}^kf_d^{\sum\limits_{d|T}\mu(\frac{T}{d})\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor}=\prod_{T=1}^k\prod_{d=1}^kf_d^{\mu(\frac{T}{d})\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor}\\=\prod_{T=1}^k(\prod_{d=1}^kf_d^{\mu(\frac{T}{d})})^{\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor} \]

中间那坨可以直接 \(O(nlogn)\) 预处理,然后数论分块即可

代码

#include<iostream>
#include<cstdio>
#define MOD (1000000007)
using namespace std;
const int N = 1e6 + 5;
typedef long long ll;

bool st[N];
int prime[N], mu[N], tot, n, m, g[N], f[N];

int qmi(int a, int b)
{
    int res = 1;
    while (b)
    {
        if (b & 1)
            res = (ll)res * a % MOD;
        a = (ll)a * a % MOD;
        b >>= 1;
    }
    return res;
}

void init()
{
    g[0] = f[1] = mu[1] = g[1] = 1;
    for (int i = 2; i < N; i++)
    {
        g[i] = 1;
        f[i] = ((ll)f[i - 1] + f[i - 2]) % MOD;
        if (!st[i])
            prime[++tot] = i, mu[i] = -1;
        for (int j = 1; i * prime[j] < N; j++)
        {
            st[i * prime[j]] = true;
            if (i % prime[j] == 0)
                break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i < N; i++)
    {
        for (int j = i; j < N; j += i)
        {
            if (mu[j / i] == 1)
                g[j] = ((ll)g[j] * f[i]) % MOD;
            else if (mu[j / i] == -1)
                g[j] = ((ll)g[j] * qmi(f[i], MOD - 2)) % MOD;
        }
    }
    for (int i = 1; i < N; i++)
        g[i] = (ll)g[i] * g[i - 1] % MOD;
}

int main()
{
    init();
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        int k = min(n, m);
        int ans = 1;
        for (int l = 1, r; l <= k; l = r + 1)
        {
            r = min(n / (n / l), m / (m / l));
            int t = (ll)(n / l) * (m / l) % (MOD - 1);
            ans = ((ll)ans * qmi((ll)g[r] * qmi(g[l - 1], MOD - 2) % MOD, t)) % MOD;
        }
        printf("%d\n", ((ll)ans + MOD) % MOD);
    }
    return 0;
}