BZOJ3992: [SDOI2015]序列统计(NTT 原根 生成函数)

题意

题目链接

给出大小为\(s\)的集合，从中选出\(n\)个数，满足他们的乘积\(\% m = x\)的方案数

sol

神仙题orz

首先不难列出最裸的dp方程，设\(f[i][j]\)表示选了\(i\)个数，他们的乘积为\(j\)的方案数

设\(g[k] = [\exists a_i = k]\)

转移的时候

\[f[i + 1][(j * k) \% m] += f[i][j] * g[k]\]

不难发现每次的转移都是相同的，因此可以直接矩阵快速幂，时间复杂度变为\(logn m^2\)

观察上面的式子，如果我们能把\((j * k) \% m\)，变成\((j + k) \% m\)的话，就是一个循环卷积的形式了

这里可以用原根来实现，设\(g\)表示\(m\)的原根，\(mp[i] = j\)表示\(g^i = j\)

直接对每个物品构造生成函数，利用mp转移即可

因为转移是个循环卷积，所以统计答案的时候应该把第\(i\)项和第\(i+m-1\)项的系数加起来

至于为啥只统计一项。

#include<bits/stdc++.h>
using namespace std;
const int mod = 1004535809, g = 3, gi = 334845270, maxn = 1e5 + 10; 
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, x, s;
int r[maxn], lim, l, ind[maxn], s[maxn], f[maxn], a[maxn], b[maxn];
int mul(int a, int b) {
    return 1ll * a * b % mod;
}
int add(int x, int y) {
    if(x + y < 0) return x + y + mod;
    return x + y >= mod ? x + y - mod : x + y;
}
int dec(int x, int y) {
    return x - y < 0 ? x - y + mod : x - y;
}
int fp(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) {
    static int q[maxn]; int tot = 0, tp = x - 1;
    for(int i = 2; i * i <= tp; 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(fp(i, (x - 1) / q[j], x) == 1) break;
        if(j == tot + 1) return i;
    }
}
void ntt(int *a, int n, int type) {
    for(int i = 1; i < n; i++) if(i < r[i]) swap(a[i], a[r[i]]);
    for(int mid = 1; mid < n; mid <<= 1) {
        int r = mid << 1, wn = fp(type == 1 ? g : gi, (mod - 1) / r, mod);
        for(int j = 0; j < lim; j += r) {
            for(int w = 1, k = 0; k < mid; k++, w = mul(w, wn)) {
                int x = a[j + k], y = mul(w, a[j + k + mid]);
                a[j + k] = add(x, y);
                a[j + k + mid] = dec(x, y);
            }
        }
    }
    if(type == -1) {
        for(int i = 0, inv = fp(lim, mod - 2, mod); i < n; i++) a[i] = mul(a[i], inv);
    }
}
void mul(int *a1, int *b1, int *c) {
    memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));//tag
    for(int i = 0; i < m - 1; i++) a[i] = a1[i], b[i] = b1[i];
    ntt(a, lim, 1); ntt(b, lim, 1);
    for(int i = 0; i < lim; i++) a[i] = mul(a[i], b[i]);
    ntt(a, lim, -1);
    for(int i = 0; i < m - 1; i++) c[i] = add(a[i], a[i + m - 1]);
}
void pre() {
    lim = 1;
    while(lim <= 2 * (m - 2)) lim <<= 1, l++;
    for(int i = 0; i < lim; i++) r[i] = (r[i >> 1] >> 1) | (i & 1) << (l - 1);
    int d = getg(m);
    for(int i = 0; i < m - 1; i++) ind[fp(d, i, m)] = i;
}
int main() {
    n = read(); m = read(); x = read(); s = read();
    pre();
    for(int i = 1; i <= s; i++) {
        int x = read();
        if(x) f[ind[x]]++;
    }
    s[ind[1]] = 1;
    while(n) {
        if(n & 1) mul(s, f, s);
        mul(f, f, f); n >>= 1;
    }
    printf("%d", s[ind[x]]);
    return 0;
}
/*
40000000 3 1 2
1 2

4 3 1 2
1 2
*/