[SDOI2017]遗忘的集合

标签：表示   展开   ref   vector   集合   stat   data   algorithm   ++   

题面

题解

设\(a_i = 0/1\)表示元素\(i\)是否在集合\(S\)中。

那么\(f\)的生成函数为\(\displaystyle F(x) = \prod_{i=1}^\infty \left(\frac 1{1 - x ^ i}\right) ^ {a_i}\)，于是问题就变成了我们已知\(F\)，求\(a\)。

两边同时取负对数，得到\(\displaystyle -\ln F(x) = \sum_{i=1}^\infty a_i \ln (1 - x ^ i)\)，对\(\ln (1 - x ^ i)\)进行泰勒展开，有

\(\displaystyle -\ln F(x) = \sum_{i = 1} ^ \infty a_i \sum_{j=1} ^ \infty -\frac{x ^ {ij}} j\)，将\(ij\)换元为\(T\)，有\(\displaystyle \ln F(x) = \sum_{T=1}^\infty x ^ T \sum_{i | T} a_i \times \frac iT\)。

我们对\(F(x)\)求一个\(\ln\)，那么\(\frac 1T\sum_{i|T} a_i \times i = [x ^ T](\ln F(x))\)，莫比乌斯反演一下\(a_i\)就出来了。

代码

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)

inline int read()
{
    int data = 0, w = 1; char ch = getchar();
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
    return data * w;
}

const int maxn(6e5 + 10);
const double pi(acos(-1));
struct complex { double x, y; } w[maxn];
inline complex conj(const complex &x) { return (complex) {x.x, -x.y}; }
inline complex operator + (const complex &lhs, const complex &rhs)
    { return (complex) {lhs.x + rhs.x, lhs.y + rhs.y}; }
inline complex operator - (const complex &lhs, const complex &rhs)
    { return (complex) {lhs.x - rhs.x, lhs.y - rhs.y}; }
inline complex operator * (const complex &lhs, const complex &rhs)
{
    return (complex) {lhs.x * rhs.x - lhs.y * rhs.y,
        lhs.x * rhs.y + lhs.y * rhs.x};
}

int r[maxn], Mod;
int fastpow(int x, int y)
{
    int ans = 1;
    for (; y; y >>= 1, x = 1ll * x * x % Mod)
        if (y & 1) ans = 1ll * ans * x % Mod;
    return ans;
}

void FFT(complex *p, int N)
{
    for (int i = 0; i < N; i++) if (i < r[i]) std::swap(p[i], p[r[i]]);
    for (int i = 1, s = 2, t = N; i < N; i <<= 1, s <<= 1, t >>= 1)
        for (int j = 0; j < N; j += s) for (int k = 0, o = 0; k < i; ++k, o += t)
        {
            complex x = p[j + k], y = w[o] * p[i + j + k];
            p[j + k] = x + y, p[i + j + k] = x - y;
        }
}

void MTT(int N, int *lhs, int *rhs, int *ans)
{
    int i; static complex a[maxn], b[maxn]; typedef long long Long;
    static complex A[maxn], B[maxn], C[maxn], D[maxn];
    for (i = 0; i < N; i++) a[i].x = lhs[i] & 0x7fff, a[i].y = lhs[i] >> 15;
    for (i = 0; i < N; i++) b[i].x = rhs[i] & 0x7fff, b[i].y = rhs[i] >> 15;
    FFT(a, N), FFT(b, N);
    for (i = 0; i < N; i++)
    {
        int j = (N - i) & (N - 1);
        static complex da, db, dc, dd;
        da = (a[i] + conj(a[j])) * (complex) {.5, 0.};
        db = (a[i] - conj(a[j])) * (complex) {0, -.5};
        dc = (b[i] + conj(b[j])) * (complex) {.5, 0.};
        dd = (b[i] - conj(b[j])) * (complex) {0, -.5};
        A[j] = da * dc, B[j] = da * dd, C[j] = db * dc, D[j] = db * dd;
    }
    for (i = 0; i < N; i++) a[i] = A[i] + B[i] * (complex) {0, 1};
    for (i = 0; i < N; i++) b[i] = C[i] + D[i] * (complex) {0, 1};
    FFT(a, N), FFT(b, N);
    for (i = 0; i < N; i++)
    {
        int da = (Long) (a[i].x / N + .5) % Mod,
            db = (Long) (a[i].y / N + .5) % Mod,
            dc = (Long) (b[i].x / N + .5) % Mod,
            dd = (Long) (b[i].y / N + .5) % Mod;
        ans[i] = (da + ((Long) (db + dc) << 15) % Mod + ((Long) dd << 30) % Mod) % Mod;
    }
}

void Inv(int *a, int *b, int N)
{
    static int c[maxn];
    if (N == 1) return (void) (*b = fastpow(*a, Mod - 2));
    Inv(a, b, (N + 1) >> 1); int len = 1, P = -1;
    while (len < (N << 1)) len <<= 1, ++P;
    for (int i = 0; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << P);
    for (int i = 0; i < len; i++) w[i] = (complex) {cos(pi * i / len), sin(pi * i / len)};
    std::copy(a, a + N, c), std::fill(c + N, c + len, 0);
    MTT(len, c, b, c);
    std::fill(c + N, c + len, 0);
    MTT(len, c, b, c);
    for (int i = 0; i < N; i++) b[i] = (2ll * b[i] - c[i] + Mod) % Mod;
}

void Ln(int *f, int *g, int N)
{
    static int A[maxn], B[maxn];
    for (int i = 1; i < N; i++) A[i - 1] = 1ll * i * f[i] % Mod;
    A[N - 1] = 0, Inv(f, B, N); int P = -1, len = 1;
    while (len < (N << 1)) len <<= 1, ++P;
    for (int i = 0; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << P);
    for (int i = 0; i < len; i++) w[i] = (complex) {cos(pi * i / len), sin(pi * i / len)};
    MTT(len, A, B, A), g[0] = 0;
    for (int i = 1; i < N; i++) g[i] = 1ll * A[i - 1] * fastpow(i, Mod - 2) % Mod;
}

int n, f[maxn], g[maxn], ans;
int main()
{
    n = read(), Mod = read(), f[0] = 1;
    for (int i = 1; i <= n; i++) f[i] = read();
    Ln(f, g, n + 1);
    for (int i = 1; i <= n; i++) g[i] = 1ll * g[i] * i % Mod;
    for (int i = 1; i <= n; i++)
        for (int j = i + i; j <= n; j += i)
            g[j] = (g[j] - g[i] + Mod) % Mod;
    for (int i = 1; i <= n; i++) if (g[i]) ++ans;
    printf("%d\n", ans);
    for (int i = 1; i <= n; i++) if (g[i]) printf("%d ", i);
    return 0;
}

[SDOI2017]遗忘的集合

标签：表示   展开   ref   vector   集合   stat   data   algorithm   ++   

原文地址：https://www.cnblogs.com/cj-xxz/p/10977520.html