题意：设$F_i$为斐波拉契数列，求

$$\sum _{i=0}^N(F_{iC}{)}^k$$

模$10^9+9$

$N,C\le 10^{18},k\le 10^5$

把斐波拉契暴力拆开

$$F_{iC}=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2}{)}^{iC}-(\frac{1-\sqrt{5}}{2}{)}^{iC}]$$

为了方便，忽略常数写成

$$F_{ic}=A^i-B^i$$

所以

$$ans=\sum _{i=0}^N(A^i-B^i{)}^k$$

$$=\sum _{i=0}^N\sum _{j=0}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)(-1{)}^j{A}^{(k-j)i}{B}^j$$

$$=\sum _{i=0}^k(-1{)}^i(\genfrac{}{}{0ex}{}{k}{j})\sum _{j=0}^N(A^{k-j}{B}^j{)}^i$$

等比数列求和即可

复杂度$O(n\log M)$

推式子先从简单的入手

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>
using namespace std;
#define MAXN 100005
typedef long long ll;
const int MOD=1e9+9,N=1e5,p=383008016,A=(p+1ll)*(MOD+1)/2%MOD,B=(1ll+MOD-p)*(MOD+1)/2%MOD;
inline int qpow(int a,int p)
{int ans=1;while (p){if (p&1) ans=(ll)ans*a%MOD;a=(ll)a*a%MOD;p>>=1;}return ans;
}
int fac[MAXN],finv[MAXN],px[MAXN],py[MAXN];
int main()
{fac[0]=1;for (int i=1;i<=N;i++) fac[i]=(ll)fac[i-1]*i%MOD;finv[N]=qpow(fac[N],MOD-2);for (int i=N-1;i>=0;i--) finv[i]=(ll)finv[i+1]*(i+1)%MOD;int T;scanf("%d",&T);while (T--){ll n,c;int k;scanf("%lld%lld%d",&n,&c,&k);c%=MOD-1;int ans=0;int x=qpow(A,c),y=qpow(B,c);px[0]=py[0]=1;for (int i=1;i<=k;i++) px[i]=(ll)px[i-1]*x%MOD,py[i]=(ll)py[i-1]*y%MOD;for (int i=0;i<=k;i++) {int t=((i&1)? MOD-1ll:1ll)*fac[k]%MOD*finv[i]%MOD*finv[k-i]%MOD;int a=(ll)px[k-i]*py[i]%MOD;if (a==1) ans=(ans+(n+1ll)%MOD*t)%MOD;else ans=(ans+t*(qpow(a,(n+1ll)%(MOD-1))-1ll)%MOD*qpow(a-1,MOD-2))%MOD;}ans=(ll)ans*qpow(p,MOD-1-k)%MOD;printf("%d\n",ans);}return 0;
}