子集卷积

发表于 | 阅读数

$O(3^n)$就是简单枚举子集。

设$p[i]=i的二进制个数$，

想成一个矩阵，也就是$c_{p(m)}$也就是$a_{p(i)},b_{p(j)}$这两行$FWT$或一下。然后我们枚举行即可。复杂度$O(n^22^n)$

P6097 【模板】子集卷积 
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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 = 1 << 20;
const int M = 1e5 + 10;
const int mod = 1e9 + 9;
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 x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}
int inc(int x, int y, int mo)
{
    if (y < 0)
        y += mo;
    if (x + y >= mo)
        x -= mo;
    return x + y;
}
int Inv2;
void FWT(int *A, int n, int op, int t) //t=1 or t=2 and t=3 xor
{
    for (int i = 2; i <= n; i <<= 1)
    {
        for (int j = 0, mid = i >> 1; j < n; j += i)
            for (int k = 0; k < mid; k++)
            {

                if (t == 1)
                    A[j + mid + k] = inc(A[j + mid + k], A[j + k] * op, mod);
                else if (t == 2)
                    A[j + k] = inc(A[j + k], A[j + mid + k] * op, mod);
                else if (t == 3)
                {
                    int x = A[j + k], y = A[j + mid + k];
                    if (op == 1)
                        A[j + k] = (x + y) % mod, A[j + mid + k] = (x - y + mod) % mod;
                    else
                        A[j + k] = 1ll * Inv2 * (x + y) % mod, A[j + mid + k] = 1ll * Inv2 * (x - y + mod) % mod;
                }
            }
    }
}

int f[21][N], g[21][N];
int res[21][N];

int main()
{
    int n = read();
    int M = 1 << n;
    for (int i = 0; i < M; i++)
        f[__builtin_popcount(i)][i] = read();
    for (int i = 0; i < M; i++)
        g[__builtin_popcount(i)][i] = read();
    for (int i = 0; i <= n; i++)
        FWT(f[i], M, 1, 1), FWT(g[i], M, 1, 1);

    for (int x = 0; x <= n; x++)
    {
        for (int i = 0; i <= x; i++)
            for (int j = 0; j < M; j++)
                res[x][j] = inc(res[x][j], 1ll * f[i][j] * g[x - i][j] % mod, mod);
        FWT(res[x], M, -1, 1);
    }
    for (int i = 0; i < M; i++)
        printf("%d ", res[__builtin_popcount(i)][i]);
}