Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

子集卷积模板

快速子集卷积。

查看代码
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#include <cstdio>
using namespace std;
typedef long long LL;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 21, mod = 1e9 + 9;
int cnt[1 << N];
void fwt(int *x, int bit, int op)
{
int tot = 1 << bit;
for (int mid = 1; mid < tot; mid <<= 1)
for (int i = 0; i < tot; i += mid << 1)
for (int j = 0; j < mid; j++)
(x[i + j + mid] += x[i + j] * op) %= mod;
}
int main()
{
static int bit, A[N][1 << N], B[N][1 << N], C[N][1 << N];
read(bit);
for (int i = 0; i < 1 << bit; i++)
cnt[i] = cnt[i >> 1] + (i & 1);
for (int i = 0; i < 1 << bit; i++)
read(A[cnt[i]][i]);
for (int i = 0; i < 1 << bit; i++)
read(B[cnt[i]][i]);
for (int i = 0; i <= bit; i++)
fwt(A[i], bit, 1), fwt(B[i], bit, 1);
for (int i = 0; i <= bit; i++)
for (int j = 0; i + j <= bit; j++)
for (int k = 0; k < 1 << bit; k++)
(C[i + j][k] += (LL)A[i][k] * B[j][k] % mod) %= mod;
for (int i = 0; i <= bit; i++)
fwt(C[i], bit, -1);
for (int i = 0; i < 1 << bit; i++)
write((C[cnt[i]][i] + mod) % mod), putchar(' ');
return 0;
}