P3321 [SDOI2015]序列统计

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给定整数 $x$，求所有可以生成出的数列，长度为$n$,且满足数列中所有数的乘积 $\mod m$ 的值等于 $x$ 的不同的数列的有多少个。

如果是加法没有$\%$就是$S$卷起来$n次$。
$f[i][j]$，表示$i$个元素乘积$\mod m=x$的方案数

为了将变成卷积的形式需要将乘法变成加法,，取个$log$即可，但如何处理$\mod m$，就需要知道$log_p x\%m$

假设$=i$

此时需要$p^i\%m\rightarrow i$一一对应，取原根即可。$g^1….g^{m-1},或者g^0….g^{m-2}$

转换完之后$f[1]$卷上$n$次，多项式快速幂。

  • 本题中$0$无意义，就不存在$\mod m=0$的情况，$g^0与g^{m-1}$都是$1$,为了保证一一对应$log_g 1\%m=1$。
  • 每次NNT需要吧多余的项加到$\%{m-1}，g^i=g^{i+m-1}$上,如果取$g^{m-1}$，会有些麻烦。
  • 注意数组清空
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 6e4 + 10;
const int mod = 1004535809;

int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int qpow(int a, int p, int mod)
{
    int base = 1;
    while (p)
    {
        if (p & 1)
            base = 1ll * base * a % mod;
        a = 1ll * a * a % mod;
        p >>= 1;
    }
    return base;
}
int GetG(int x)
{
    int q[N];
    int tot = 0, tp = x - 1;
    for (int i = 2; i * i <= tp; i++)
    { //这里是i * i
        if (!(tp % i))
        {
            q[++tot] = i;
            while (!(tp % i))
                tp /= i;
        }
    }
    if (tp > 1)
        q[++tot] = tp;
    for (int i = 2, j; i <= x - 1; i++)
    {
        for (j = 1; j <= tot; j++)
            if (qpow(i, (x - 1) / q[j], x) == 1)
                break;
        if (j == tot + 1)
            return i;
    }
}
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}

const int P = 1004535809, gi = 3;
const double pi = acos(-1.0);
int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}
int A[N], B[N], res[N];
void NTTX(int *a, int n, int *b, int m, int *ans, int mod_x)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    for (int i = 0; i <= ML; i++)
        A[i] = a[i];
    for (int i = 0; i <= ML; i++)
        B[i] = b[i];
    NTT(A, ML, 1);
    NTT(B, ML, 1);
    for (int i = 0; i < ML; i++)
        A[i] = 1ll * A[i] * B[i] % P;
    NTT(A, ML, -1);
    for (int i = 0; i < n; i++)
        res[i] = (A[i] + A[i + n]) % mod;

    for (int i = 0; i < n; i++)
        ans[i] = res[i];
}
int mp[N];
void init(int m)
{
    int gg = GetG(m);
    int tmp = 1;
    for (int i = 0; i < m - 1; i++)
    {
        mp[tmp] = i;
        tmp = 1ll * tmp * gg % m;
    }
}
int ans[N], a[N];
int main()
{
    int n = read(), m = read(), x = read(), s = read();
    init(m);
    for (int i = 1; i <= s; i++)
    {
        int p = read();
        if (p)
            a[mp[p]]++;
    }
    // NTTX(a, m - 1, a, m - 1, a, m - 1);
    // NTTX(a, m - 1, a, m - 1, a, m - 1);
    // for (int i = 0; i <= 10; i++)
    //     cout << a[i] << " ";
    ans[mp[1]] = 1;
    while (n)
    {
        if (n & 1)
            NTTX(ans, m - 1, a, m - 1, ans, m - 1);
        NTTX(a, m - 1, a, m - 1, a, m - 1);
        n >>= 1;
    }

    printf("%d\n", ans[mp[x]]);
}