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快速乘法逆元（在线）

用法

在线计算 \(x = [x_1, x_2, \cdots, x_n]\) 在模 \(p\) 意义下的逆元。

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#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e7 + 3;
pair<int, int> F[MAXN], G[MAXN];
int I[MAXN];
using u32 = uint32_t;
u32 read(u32 &seed){
    seed ^= seed << 13;
    seed ^= seed >> 17;
    seed ^= seed << 5;
    return seed;
}
int main(){
    ios :: sync_with_stdio(false);
    cin.tie(nullptr);
    u32 seed;
    int n, p;
    cin >> n >> p >> seed;
    int m = pow(p, 1.0 / 3.0);
    I[1] = 1;
    for(int i = 2;i <= p / m;++ i){
        I[i] = 1ll * (p / i) * (p - I[p % i]) % p;
    }
    for(int i = 1;i < m;++ i){
        for(int j = i + 1;j <= m;++ j){
            if(!F[i * m * m / j].second){
                F[i * m * m / j] = { i, j };
                G[i * m * m / j] = { i, j };
            }
        }
    }
    F[    0] = G[    0] = { 0, 1 };
    F[m * m] = G[m * m] = { 1, 1 };
    for(int i = 1;i <      m * m;++ i) if(!F[i].second)
        F[i] = F[i - 1];
    for(int i = m * m - 1;i >= 1;-- i) if(!G[i].second)
        G[i] = G[i + 1];
    int lastans = 0;
    for(int i = 1;i <= n;++ i){
        int a, inv;
        a = (read(seed) ^ lastans) % (p - 1) + 1;
        int w = 1ll * a * m * m / p;
        auto &yy1 = F[w].second;    // *avoid y1 in <cmath>
        if(1ll * a * yy1 % p <= p / m){
            inv = 1ll * I[1ll * a * yy1 % p] * yy1 % p;
        } else {
            auto &yy2 = G[w].second;
            inv = 1ll * I[1ll * a * (p - yy2) % p] * (p - yy2) % p;
        }
        lastans = inv;
    }
    cout << lastans << "\n";
    return 0;
}