单位根反演

之前碰到过两次但是没学

https://acm.hdu.edu.cn/showproblem.php?pid=7013

hdu多校的题

单位根反演都比较套路

如果有组合数可以把常系数和它化成n次幂的形式

实现复杂度从$n/k$变到$k$

详细的有空补

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

//#include <immintrin.h>
//#include <emmintrin.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (int i=h;i<=t;i++)
#define dep(i,t,h) for (int i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rint register int
inline ll rd(){
    ll x=0;char c=getchar();bool f=0;
    while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
char ss[1<<24],*A=ss,*B=ss;
IL char gc()
{
    return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T>void maxa(T &x,T y)
{
    if (y>x) x=y;
}
template<class T>void mina(T &x,T y)
{
    if (y<x) x=y;
}
template<class T>void read(T &x)
{
    int f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
    while(c=gc(),c>47&&c<58) x=x*10+(c^48); x*=f;
}
const int mo=1e9+9;
ll fsp(ll x,ll y)
{
    if (y==1) return x;
    ll ans=fsp(x,y/2);
    ans=ans*ans%mo;
    if (y%2==1) ans=ans*x%mo;
    return ans;
}
struct cp {
    ll x,y;
    cp operator +(cp B)
    {
        return (cp){x+B.x,y+B.y};
    }
    cp operator -(cp B)
    {
        return (cp){x-B.x,y-B.y};
    }
    ll operator *(cp B)
    {
        return x*B.y-y*B.x;
    }
    int half() { return y < 0 || (y == 0 && x < 0); }
};
struct re{
    int a,b,c;
};
const int g=13;
const int N=5e5;
ll k,l,n;
ll f1[N],f2[N];
ll h[510][510],p[510][510];
struct re2{
    ll a[510][510];
    void operator *=(const re2 b){
        re2 c;
        me(c.a);
        rep(i,0,n-1)
          rep(j,0,n-1)
            rep(k,0,n-1)
              c.a[i][k]=(c.a[i][k]+a[i][j]*b.a[j][k])%mo;
        rep(i,0,n-1)
          rep(j,0,n-1)
            a[i][j]=c.a[i][j];
    }
}a,b,c;
int main()
{
   freopen("1.in","r",stdin);
   ios::sync_with_stdio(false);
   int T;
   cin>>T;
   while (T--)
   {
        cin>>k>>l>>n;
        k=(k-2)%mo;
        int w=fsp(g,(mo-1)/n);
        f1[0]=1;
        rep(i,1,n) f1[i]=f1[i-1]*w%mo;
        int w1=fsp(w,mo-2);
        f2[0]=1;
        rep(i,1,n*n) f2[i]=f2[i-1]*w1%mo;
        rep(i,0,n-1)
          rep(j,0,n-1)
            a.a[i][j]=c.a[i][j]=f2[i*j];
        rep(i,0,n-1)
       rep(j,0,n-1)
         b.a[i][j]=fsp((f1[i]+f1[j]+k)%mo,l)*fsp(n*n,mo-2)%mo; 
     a*=b;
     a*=c;
     rep(i,0,n-1)
     {
       rep(j,0,n-2)
         cout<<a.a[i][j]<<" ";
       cout<<a.a[i][n-1]<<endl;
     }
   }
   return 0;
}