所以其实解是唯一的……
所以字典序什么的根本不存在……
迷惑行为……
令 \(a_i = [i \in S]\)，则 \(f\) 的生成函数 \[ F(x) = \sum\limits_{i=0}^{\infty} f(i)x^i = \prod\limits_{i=1}^n \left(\frac1{1-x^i}\right)^{a_i} \]
两边取 \(\ln\)，得 \[ \ln F(x) = -\sum\limits_{i=1}^n a_i\ln(1-x^i) \]
对右边泰勒展开，得 \[ \ln F(x) = \sum\limits_{i=1}^n a_i \sum\limits_{j=1}^{\infty} \frac{x^{ij}}j \]
根据某些莫反套路，换元，令 \(T=ij\)，得 \[ \ln F(x) = \sum\limits_{T=1}^{\infty} \frac{x^T}T \sum\limits_{d|T} a_dd \]
听说要莫反，然而其实直接枚举倍数减掉贡献就行了。
然后就能求出 \(a_i\) 了。
然后这题就切了。
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using namespace std;
const int N = 1 << 19;
const double pi = acos(-1);
const int W = 1 << 15;
int len,mod,n,ans;
int lg2[N + 5],rev[N + 5];
int fac[N + 5],ifac[N + 5],inv[N + 5];
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
struct poly
{
int a[N + 5];
inline const int &operator[](int x) const
{
return a[x];
}
inline int &operator[](int x)
{
return a[x];
}
inline void clear(int x = 0)
{
memset(a + x,0,(N - x + 1) << 2);
}
} f;
struct cp
{
double a,b;
inline void operator+=(const cp &o)
{
a += o.a,b += o.b;
}
inline cp operator+(const cp &o) const
{
return (cp){a + o.a,b + o.b};
}
inline cp operator-(const cp &o) const
{
return (cp){a - o.a,b - o.b};
}
inline cp operator*(const cp &o) const
{
return (cp){a * o.a - b * o.b,a * o.b + b * o.a};
}
inline void operator*=(const cp &o)
{
*this = *this * o;
}
inline cp operator*(const double &o) const
{
return (cp){a * o,b * o};
}
inline cp operator~() const
{
return (cp){a,-b};
}
} rt[N + 5];
inline void init(int len)
{
for(n = 1;n < len;n <<= 1);
for(register int i = 2;i <= n;++i)
lg2[i] = lg2[i >> 1] + 1;
for(register int i = 0;i <= (n >> 1);++i)
rt[(n >> 1) + i] = (cp){cos(2 * pi * i / n),sin(2 * pi * i / n)};
for(register int i = (n >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= n;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[n] = fpow(fac[n],mod - 2);
for(register int i = n;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= n;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
inline void fft(cp *a,int type,int n)
{
type == -1 && (reverse(a + 1,a + n),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
cp t = rt[m | j] * a[i | j | m];
a[i | j | m] = a[i | j] - t,a[i | j] += t;
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i].a /= n,a[i].b /= n;
}
inline void mul(poly &a,const poly &b,int n)
{
static cp f[N + 5],g[N + 5],h[N + 5];
int lim = 1;
memset(f,0,sizeof f),memset(g,0,sizeof g);
for(;lim < (n << 1);lim <<= 1);
for(register int i = 0;i < n;++i)
f[i] = (cp){a[i] / W,a[i] % W},g[i] = (cp){b[i] / W,b[i] % W};
fft(f,1,lim),fft(g,1,lim);
for(register int i = 0;i < lim;++i)
h[i] = ~f[(lim - i) % lim];
for(register int i = 0;i < lim;++i)
f[i] *= g[i],g[i] *= h[i];
fft(f,-1,lim),fft(g,-1,lim);
for(register int i = 0;i < lim;++i)
{
long long ac = (f[i].a + g[i].a) / 2 + 0.5;
long long bd = g[i].a - ac + 0.5;
long long bcad = f[i].b + 0.5;
a[i] = ((ac % mod * W % mod * W % mod) % mod + (bcad % mod * W % mod) % mod + bd % mod) % mod;
}
}
inline poly inverse(const poly &f,int n)
{
static int s[30];
static poly g,h,q;
int top = 0;
g.clear();
for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
g[0] = fpow(f[0],mod - 2);
for(;top;--top)
{
n = s[top];
q = g,h = f,h.clear(n);
mul(g,g,n),g.clear(n),mul(g,h,n);
for(register int i = 0;i < n;++i)
g[i] = dec(add(q[i],q[i]),g[i]);
g.clear(n);
}
return g;
}
inline void derivative(poly &f,int n)
{
for(register int i = 1;i < n;++i)
f[i - 1] = (long long)f[i] * i % mod;
f[n - 1] = 0;
}
inline void integral(poly &f,int n)
{
for(register int i = n - 1;~i;--i)
f[i + 1] = (long long)f[i] * inv[i + 1] % mod;
f[0] = 0;
}
inline poly ln(const poly &f,int n)
{
static poly g;
g = f,derivative(g,n),mul(g,inverse(f,n),n),integral(g,n);
return g;
}
int main()
{
scanf("%d%d",&len,&mod),init((len + 1) << 1),f[0] = 1;
for(register int i = 1;i <= len;++i)
scanf("%d",f.a + i);
f = ln(f,len + 1);
for(register int i = 1;i <= len;++i)
f[i] = (long long)f[i] * i % mod;
for(register int i = 1;i <= len;++i)
{
ans += (bool)f[i];
for(register int j = 2 * i;j <= len;j += i)
f[j] = dec(f[j],f[i]);
}
printf("%d\n",ans);
for(register int i = 1;i <= len;++i)
f[i] && (printf("%d ",i),1);
}