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2019牛客第五场C题(generator2) BSGS

2019牛客第五场C题(generator2) BSGS

作者: 叔丁基锂_ | 来源:发表于2019-08-02 12:00 被阅读0次

    题意:给出n,x_0a,b,p(1≤n≤10^{18},\,0≤x_p,a,b<p≤10^9+9,p\text{ is a prime number}).已知序列x 满足x_i=(ax_{i-1}+b) \bmod p ,再给出Q(Q \le 1000)组询问每次一个数v ,求最小的index使得x_{index}=v , 不存在则输出-1

    题解:令y_i=(x_i+b\cdot(a-1)^{-1}) \bmod p, 于是有y_i=ay_{i-1} \bmod p, 于是yi就成了一个等比数列。这样我们就发现本题适用于BSGS算法。注意到传统的BSGS算法都是分成\Theta(\sqrt{p}) 的两个部分,但是由于这个题的Q非常大,于是我们可以预处理(即baby-step部分)更多的项:例如,p^{2/3} 。最后特判一下a=0和a=1的情况,这个题就做完了。

    一些玄学的情况:这份代码ac过(虽然时间有点勉强),但是后来交是tle的,再等一会又能ac了(牛客的测评机啊。。。。)

    #include <algorithm>
    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <unordered_map>
    #include <vector>
    #define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
    using namespace std;
    using ll = long long;
    const ll maxn = 1e6;
    ll mod;
      
    ll bin(ll x, ll n, ll MOD)
    {
        ll ret = MOD != 1;
        for (x %= MOD; n; n >>= 1, x = x * x % MOD)
            if (n & 1)
                ret = ret * x % MOD;
        return ret;
    }
    inline ll get_inv(ll x, ll p) { return bin(x, p - 2, p); }
      
    unordered_map<ll, ll> mp;
    ll m, ma;
    ll init(ll a, ll p)
    {
        mp.clear();
        ll v = 1;
        m = pow(p + 1.5, 2.0 / 3.0);
        ma = pow(p + 1.5, 1.0 / 3.0) + 3;
        FOR(i, 1, m + 1)
        {
            v = v * a % p;
            mp[v] = i;
        }
        return v;
    }
      
    ll BSGS(ll a, ll b, ll p, ll init_d)
    { // a^x = b (mod p)
        a %= p;
        if (!a && !b)
            return 1;
        if (!a)
            return -1;
        ll v = init_d;
        ll vv = init_d;
        ll inv_b = get_inv(b, p);
      
        FOR(i, 1, ma + 1)
        {
            auto it = mp.find(vv * inv_b % p);
            if (it != mp.end())
                return i * m - it->second;
            vv = vv * v % p;
        }
        return -1;
    }
      
    int main()
    {
        ios::sync_with_stdio(false);
        int round;
        cin >> round;
        while (round--)
        {
            ll n, x0, a, b, p;
            int Q;
            cin >> n >> x0 >> a >> b >> p;
            cin >> Q;
            if (a == 0)
            {
                while (Q--)
                {
                    ll v;
                    cin >> v;
                    if (v == x0)
                    {
                        cout << 0 << endl;
                    }
                    else if (v == b)
                    {
                        cout << 1 << endl;
                    }
                    else
                    {
                        cout << -1 << endl;
                    }
                }
            }
            else if (a == 1)
            {
                ll invb = get_inv(b, p);
                while (Q--)
                {
                    ll v;
                    cin >> v;
                    ll ans = (((v - x0 + p) % p) * invb) % p;
                    if (ans >= n)
                    {
                        cout << -1 << endl;
                    }
                    else
                    {
                        cout << ans << endl;
                    }
                }
            }
            else
            {
                ll bais = b * get_inv(a - 1, p) % p;
                ll y0 = (x0 + bais) % p;
                ll inv_y0 = get_inv(y0, p);
                // m = min(maxn, n);
                // ma = p / maxn + 3;
                ll init_d = init(a, p);
                while (Q--)
                {
                    ll v;
                    cin >> v;
                    v = (v + bais) % p;
                    v = (v * inv_y0) % p;
                    ll res = BSGS(a, v, p, init_d);
                    if (res >= n)
                        res = -1;
                    cout << res << endl;
                }
            }
        }
    }
    

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