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 = 8005;

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

namespace Poly {
    const int L = 1 << 14;
    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 = 1;
        for (; y > 0; y >>= 1) {
            if (y & 1) res = res * x % mod;
            x = x * x % mod;
        }
        return res;
    }
    
    inline void prework(int len) {
        int l = 0;
        for (lim = 1; lim < len; lim <<= 1, ++l);
        for (int i = 0; i < lim; i++)
            pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1));
    }
    
    inline void ntt(poly &c, int opt) {
        c.resize(lim, 0);
        for (int i = 0; i < lim; i++)    
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = 1; i < lim; i <<= 1) {
            ll wn = fpow(3, (P - 1) / (i << 1));
            if (opt == -1) wn = fpow(wn, P - 2);
            for (int len = i << 1, j = 0; j < lim; j += len) {
                ll w = 1;
                for (int k = 0; 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 == -1) {
            ll inv = fpow(lim, P - 2);
            for (int i = 0; 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() - 1);
        ntt(u, 1), ntt(v, 1);
        for (int i = 0; i < lim; i++) res.push_back(u[i] * v[i] % P);
        ntt(res, -1);
        res.resize(x.size() + y.size() - 1);
        return res;
    }
    
    inline void adj(poly &c) {
        for (int i = 0; i < m - 1; i++) inc(c[i], c[i + m - 1]);
        c.resize(m - 1);
    }
    
    inline poly pow(poly x, int y) {
        poly res;
        res.resize(m - 1);
        res[0] = 1;
        for (; y; y >>= 1) {
            if (y & 1) 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 = 0; char ch = 0; T op = 1;
    for (; ch > '9'|| ch < '0'; ch = getchar())
        if (ch == '-') op = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

inline int getRoot() {
    int cnt = 0;
    for (int i = 2; i <= m - 2; i++)
        if ((m - 1) % i == 0) fac[++cnt] = i;
    for (int i = 2; ; i++) {
        bool flag = 1;
        for (int j = 1; j <= cnt; j++)
            if (fpow(i, fac[j], m) == 1) {
                flag = 0;
                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 = 1; i <= m - 2; i++) buc[fpow(root, i, m)] = i;
    
    poly trans;
    trans.resize(m - 1);
    for (int x, i = 1; i <= s; i++) {
        read(x);
        if (!x) continue;
        trans[buc[x]] = 1;
    }
    poly ans = pow(trans, n);
    
    printf("%lld\n", ans[buc[tar]]);
    return 0;
}

 

来源：https://www.cnblogs.com/CzxingcHen/p/10289484.html