CF914G - Sum the Fibonacci

发表于 | 阅读数

$(s_a|s_b)\&s_c\&(s_d\oplus s_e)=2^i$

$s_a \& s_b =0$

求$\sum Fib(s_a|s_b)\&Fib(s_c)\&Fib(s_d\oplus s_e)$

  • 第一个子集卷积，
  • 第三个$FWT$
  • 这三部分每部分再乘上$Fib[i]$
  • 整体卷两次积
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 = 1 << 17;

const int mod = 1e9 + 7;
int read()
{
    int x = 0, F = 1;
    char d = getchar();
    while (d < '0' || d > '9')
    {
        if (d == '-')
            F = -1;
        d = getchar();
    }
    while (d >= '0' && d <= '9')
        x = (x << 1) + (x << 3) + d - '0', d = 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;
                }
            }
    }
}
void FWTX(int *A, int *B, int n, int t)
{
    Inv2 = qpow(2, mod - 2, mod);
    FWT(A, n, 1, t), FWT(B, n, 1, t);

    for (int i = 0; i < n; ++i)
        A[i] = 1ll * A[i] * B[i] % mod;
    FWT(A, n, -1, t);
}
int F[21][N], G[21][N];
int res[21][N];
void FJU(int *a, int *b, int n, int *d)
{

    int m = 1 << n;
    for (int i = 0; i < m; i++)
        F[__builtin_popcount(i)][i] = a[i];
    for (int i = 0; i < m; i++)
        G[__builtin_popcount(i)][i] = b[i];
    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++)
        d[i] = res[__builtin_popcount(i)][i];
}

int a[N], b[N], d[N], c[N];
int Fib[N];
int main()
{
    int n = read();
    int m = 1 << 17;
    Fib[0] = 0;
    Fib[1] = 1;
    for (int i = 2; i < m; i++)
        Fib[i] = (Fib[i - 2] + Fib[i - 1]) % mod;
    for (int i = 0; i < n; i++)
    {
        int x = read();
        a[x]++;
        b[x]++;
        c[x]++;
    }
    FJU(a, b, 17, d);
    FWTX(a, b, 1 << 17, 3);
    for (int i = 0; i < m; i++)
        d[i] = 1ll * d[i] * Fib[i] % mod;
    for (int i = 0; i < m; i++)
        a[i] = 1ll * a[i] * Fib[i] % mod;
    for (int i = 0; i < m; i++)
        c[i] = 1ll * c[i] * Fib[i] % mod;

    FWTX(d, c, 1 << 17, 2);
    FWTX(d, a, 1 << 17, 2);
    int res = 0;
    for (int i = 0; i < 17; i++)
        res = inc(res, d[1 << i], mod);
    printf("%d\n", res);
}