Luogu 3321 [SDOI2015]序列统计

BZOJ 3992

点开这道题之后才发现我对原根的理解大概只停留在$998244353$的原根是$3$……

关于原根： 点我

首先写出$dp$方程，设$f_{i, j}$表示序列长度为$i$当前所有数乘积模$m$为$j$的方案数，有转移

$$f_{i, x * y \mod m} = \sum_{y \in s} f_{i - 1, x}$$

把$x$和$y$取个对数就可以变成卷积的形式了。

然而在模意义下，我们可以用原根的$k$次方来代替原来的数，这样子就达到了取对数的效果。

注意到每一次转移形式都是相同的，我们可以用快速幂的方式来优化。

时间复杂度$O(mlogmlogn)$。

然而我的代码只有在有$c++11$的时候才是对的，哪位大佬如果知道了为什么教教我呗。

Code：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
typedef vector <ll> poly;

const int N = ;

int n, m, tar, fac[N], buc[N];

namespace Poly {
    const int L =  << ;
    const ll P = 1004535809LL;

    int lim, pos[L];

    template <typename T>
    inline void inc(T &x, T y) {
        x += y;
        if (x >= P) x -= P;
    }

    inline ll fpow(ll x, ll y, ll mod = P) {
        ll res = ;
        for (; y > ; y >>= ) {
            if (y & ) res = res * x % mod;
            x = x * x % mod;
        }
        return res;
    }

    inline void prework(int len) {
        int l = ;
        for (lim = ; lim < len; lim <<= , ++l);
        for (int i = ; i < lim; i++)
            pos[i] = (pos[i >> ] >> ) | ((i & ) << (l - ));
    }

    inline void ntt(poly &c, int opt) {
        c.resize(lim, );
        for (int i = ; i < lim; i++)
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = ; i < lim; i <<= ) {
            ll wn = fpow(, (P - ) / (i << ));
            if (opt == -) wn = fpow(wn, P - );
            for (int len = i << , j = ; j < lim; j += len) {
                ll w = ;
                for (int k = ; k < i; k++, w = w * wn % P) {
                    ll x = c[j + k], y = w * c[j + k + i] % P;
                    c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P;
                }
            }
        }

        if (opt == -) {
            ll inv = fpow(lim, P - );
            for (int i = ; i < lim; i++) c[i] = c[i] * inv % P;
        }
    }

    poly operator * (const poly x, const poly y) {
        poly u = x, v = y, res;
        prework(x.size() + y.size() - );
        ntt(u, ), ntt(v, );
        for (int i = ; i < lim; i++) res.push_back(u[i] * v[i] % P);
        ntt(res, -);
        res.resize(x.size() + y.size() - );
        return res;
    }

    inline void adj(poly &c) {
        for (int i = ; i < m - ; i++) inc(c[i], c[i + m - ]);
        c.resize(m - );
    }

    inline poly pow(poly x, int y) {
        poly res;
        res.resize(m - );
        res[] = ;
        for (; y; y >>= ) {
            if (y & ) res = res * x, adj(res);
            x = x * x, adj(x);
        }
        return res;
    }

}

using Poly :: P;
using Poly :: fpow;
using Poly :: pow;

template <typename T>
inline void read(T &X) {
    X = ; char ch = ; T op = ;
    for (; ch > ''|| ch < ''; ch = getchar())
        if (ch == '-') op = -;
    for (; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}

inline int getRoot() {
    int cnt = ;
    for (int i = ; i <= m - ; i++)
        if ((m - ) % i == ) fac[++cnt] = i;
    for (int i = ; ; i++) {
        bool flag = ;
        for (int j = ; j <= cnt; j++)
            if (fpow(i, fac[j], m) == ) {
                flag = ;
                break;
            }
        if (flag) return i;
    }
}

int main() {
/*    #ifndef ONLINE_JUDGE
        freopen("8.in", "r", stdin);
    #endif   */

    int s, root;
    read(n), read(m), read(tar), read(s);

    root = getRoot();
    for (int i = ; i <= m - ; i++) buc[fpow(root, i, m)] = i;

    poly trans;
    trans.resize(m - );
    for (int x, i = ; i <= s; i++) {
        read(x);
        if (!x) continue;
        trans[buc[x]] = ;
    }
    poly ans = pow(trans, n);

    printf("%lld\n", ans[buc[tar]]);
    return ;
}