二次剩余+快速幂+离线阶乘逆元。朝鲜人的考古题，快速幂年年有，啥时候把这题改成光速幂版本，又能卡死一群。
向金日成综合大学的巨巨们低头。
1 #include<cstdio> 2 #include<cstdlib> 3 #include<algorithm> 4 using namespace std; 5 typedef long long ll; 6 7 const int P=1000000009; 8 const int INV2=500000005; 9 const int SQRT5=383008016; 10 const int INVSQRT5=276601605; 11 12 ll A0=691504013; 13 ll B0=308495997; 14 15 ll A=691504013; 16 ll B=308495997; 17 18 const int N=100005; 19 20 ll n,K; 21 ll fac[N],inv[N]; 22 ll pa[N],pb[N]; 23 24 inline void Pre(int n){ 25 fac[0]=1; for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P; 26 inv[1]=1; for (int i=2;i<=n;i++) inv[i]=(P-P/i)*inv[P%i]%P; 27 inv[0]=1; for (int i=1;i<=n;i++) inv[i]=inv[i]*inv[i-1]%P; 28 } 29 30 inline void Pre2(int n){ 31 pa[0]=1; for (int i=1;i<=n;i++) pa[i]=pa[i-1]*A%P; 32 pb[0]=1; for (int i=1;i<=n;i++) pb[i]=pb[i-1]*B%P; 33 } 34 35 inline ll C(int n,int m){ 36 return fac[n]*inv[m]%P*inv[n-m]%P; 37 } 38 39 inline ll Pow(ll a,ll b){ 40 ll ret=1; 41 for (;b;b>>=1,a=a*a%P) 42 if (b&1) 43 ret=ret*a%P; 44 return ret; 45 } 46 47 inline ll Inv(ll x){ 48 return Pow(x,P-2); 49 } 50 51 inline void Solve(){ 52 ll Ans=0; 53 for (int j=0;j<=K;j++){ 54 ll t=pa[K-j]*pb[j]%P,tem; 55 tem=t==1?n%P:t*(Pow(t,n)-1+P)%P*Inv(t-1)%P; 56 if (~j&1) 57 Ans+=C(K,j)*tem%P,Ans%=P; 58 else 59 Ans+=P-C(K,j)*tem%P,Ans%=P; 60 } 61 Ans=Ans*Pow(INVSQRT5,K)%P; 62 printf("%lld ",Ans); 63 } 64 65 int main(){ 66 int T; 67 Pre(100000); 68 scanf("%d",&T); 69 ll c; 70 while (T--){ 71 scanf("%lld%lld%lld",&n,&c,&K); 72 A=Pow(A0,c); 73 B=Pow(B0,c); 74 Pre2(K); 75 Solve(); 76 } 77 }