「SDOI2015」序列统计-DP+NTT

Decription

给定集合$S$,集合$S$中的元素都小于等于$M$.

求有多少种数列满足长度为$n$,且数列中元素乘积$\mod M$的值为$X$.

Solution

$10pts$:设$f_{i,j}$表示数列中填了$i$个数,乘积$\mod m$为$j$的方案数，每次枚举有下一元素$\mod M$的值即可.

$60pts$:发现每次的转移都是一样的,所以可以借用快速幂的思想转移,时间复杂度为$O(m^2logn)$.

$100pts:$考虑到$O(logn)$以无法优化，所以考虑优化$O(m^2)$.希望将转移中的$i*j$转化为$i+j$,这样可以使用$NTT$来优化转移.

发现$x \leqslant 1$且$m$为奇质数,所以可以用原根的指数运算代替乘法运算(对于$p$的原根$g$,$g_i(i \in [0,p-2])$完美映射了$[1,p-1]$).

所以$i*j$ 就转化为$i'+j'$,其中$g^{i'}=i,g^{j'}=j$.用$NTT$优化转移即可.

#include <bits/stdc++.h>
using namespace std;

typedef long long lint;
const int mod = 1004535809;
const int G = 3, Phi = mod - 1;
const int maxn = 33005;

int n, m, x, S, L, N;
int a[maxn], R[maxn], A[maxn], B[maxn], p[maxn], f[maxn], g[maxn];

inline int gi()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int sum = 0;
    while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
    return sum;
}

int Pow(int x, int k, int mod)
{
    int res = 1;
    while (k) {
        if (k & 1) res = (lint)res * x % mod;
        x = (lint)x * x % mod; k >>= 1;
    }
    return res;
}

int mul[maxn];
int calc_rt(int m)
{
    int phi = m - 1;
    for (int i = 2; i * i <= phi; ++i)
        if (phi % i == 0) {
            while (phi % i == 0) phi /= i;
            mul[++mul[0]] = i;
        }
    if (phi > 1) mul[++mul[0]] = phi;
    for (int i = 2; ; ++i) {
        bool flag = 0;
        for (int j = 1; j <= mul[0]; ++j)
            if (Pow(i, (m - 1) / mul[j], m) == 1) { flag = 1; break;    }
        if (!flag) return i;
    }
    return 0;
}

void prepare()
{
    N = 1;
    for (int i = (m << 1) - 2; N <= i; N <<= 1) ++L;
    for (int i = 0; i <= N; ++i) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    int rt = calc_rt(m);
    for (int i = 0; i < m - 1; ++i) p[Pow(rt, i, m)] = i;
}

void NTT(int *a, int f)
{
    for (int i = 0; i < N; ++i)
        if (i < R[i]) swap(a[i], a[R[i]]);
    for (int i = 1; i < N; i <<= 1) {
        int wn = Pow(G, Phi / (i << 1), mod), t;
        if (f == -1) wn = Pow(wn, mod - 2, mod);
        for (int j = 0; j < N; j += (i << 1)) {
            int w = 1;
            for (int k = 0; k < i; ++k, w = (lint)w * wn % mod) {
                t = (lint)a[i + j + k] * w % mod;
                a[j + i + k] = a[j + k] - t + mod; 
                if (a[j + i + k] >= mod) a[j + i + k] -= mod;
                a[j + k] = a[j + k] + t;
                if (a[j + k] >= mod) a[j + k] -= mod;
            }
        }
    }
}

void Calc(int *a, int *b)
{
    for (int i = 0; i < N; ++i) A[i] = a[i], B[i] = b[i], a[i] = 0;
    NTT(A, 1); NTT(B, 1);
    for (int i = 0; i < N; ++i) A[i] = (lint)A[i] * B[i] % mod;
    NTT(A, -1);
    int inv = Pow(N, mod - 2, mod);
    for (int i = 0; i < N; ++i) {
        (a[i % (m - 1)] += (lint)A[i] * inv % mod) %= mod;
    }
}

int main()
{
    n = gi(); m = gi(); x = gi(); S = gi();
    prepare();

    for (int a, i = 1; i <= S; ++i) {
        a = gi() % m;
        if (a) ++f[p[a]];
    }   
    g[p[1]] = 1;

    while (n) {
        if (n & 1) Calc(g, f);
        Calc(f, f); n >>= 1;
    }

    printf("%d\n", g[p[x]]);

    return 0;
}