P3704 [SDOI2017]数字表格

P3704 [SDOI2017]数字表格

链接

分析：

$\ \ \ \prod\limits_{i = 1}^{n} \prod\limits_{j = 1}^{m} f[gcd(i, j)]$

$=\prod\limits_{d = 1}^{min(n, m)} \prod\limits_{i = 1}^{n} \prod\limits_{j = 1}^{m} [gcd(i, j) = d] \times f[d]$

$=\prod\limits_{d = 1}^{min(n, m)} f[d] ^ {\sum\limits_{i = 1}^{n} \sum\limits_{j = 1}^{m} [gcd(i, j) = d]}$

$=\prod\limits_{d = 1}^{min(n, m)} f[d] ^ {\sum\limits_{k = 1}^{min( \frac{n}{d} , \frac{m}{d} )} \mu(k) \frac{n}{kd} \frac{m}{kd}}$

设$T=kd$

$\prod\limits_{T = 1} ^ {min(n, m)} (\prod\limits_{d | T} f[d] ^ {\mu(\frac{T}{d}) } ) ^ {\frac{n}{T} \frac{m}{T} }$

对中间的部分$nlogn$预处理，$O(\sqrt n)$处理每个询问。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
    for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
}

const int N = , mod = 1e9 + ;
int mu[N], pri[N], f[N], g1[N], g2[N], inv1[N], inv2[N];
bool nopri[N];

int ksm(int a,LL b) {
    register int res = ;
    while (b) {
        if (b & ) res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        b >>= ;
    }
    return res % mod;
}
void init(int n) {
    nopri[] = true; mu[] = ;
    int tot = ;
    for (int i = ; i <= n; ++i) {
        if (!nopri[i]) pri[++tot] = i, mu[i] = -;
        for (register int j = ; j <= tot && pri[j] * i <= n; ++j) {
            nopri[pri[j] * i] = ;
            if (i % pri[j] == ) { mu[i * pri[j]] = ; break; }
            mu[pri[j] * i] = -mu[i];
        }
    }
    f[] = , f[] = ; g1[] = g2[] = ;
    for (register int i = ; i <= n; ++i) f[i] = (f[i - ] + f[i - ]) % mod, g1[i] = g2[i] = ;
    for (int i = ; i <= n; ++i)
        for (int j = i; j <= n; j += i) {
            if (mu[j / i] == ) continue;
            else if (mu[j / i] == ) g1[j] = 1ll * g1[j] * f[i] % mod;
            else g2[j] = 1ll * g2[j] * f[i] % mod;
        }
    g1[] = g2[] = inv1[] = inv2[] = ;
    for (int i = ; i <= n; ++i) {
        g1[i] = 1ll * g1[i] * g1[i - ] % mod,
        g2[i] = 1ll * g2[i] * g2[i - ] % mod;
        inv1[i] = ksm(g1[i], mod - );
        inv2[i] = ksm(g2[i], mod - );
    }
}
void solve() {
    int n = read(), m = read(), nm = min(n, m), pos = , ans = ;
    for (int t1, t2, i = ; i <= nm; i = pos + ) {
        pos = min(n / (n / i), m / (m / i));
        LL t = 1ll * (n / i) * (m / i); // !!!
        t1 = 1ll * g1[pos] * inv1[i - ] % mod;
        t2 = 1ll * g2[pos] * inv2[i - ] % mod;
        ans = 1ll * ans * ksm(t1, t) % mod * ksm(ksm(t2, t), mod - ) % mod;
    }
    cout << ans << "\n";
}
int main() {
    init();
    for (int T = read(); T --; solve());
    return ;
}