题意:求$ [(\sqrt{2}+\sqrt{3})^{2n}] mod 1024 $
分析: 把指数的2带入 原式等于 $ [(5+2\sqrt{6})^n] $
有一个重要的结论是n次运算后其结果最终形式也是$ a_n+b_n\sqrt{6} $ 的形式
记最终的解$ F(n) = a_n+b_n\sqrt{6} $
$ F(n-1) = a_{n-1}+b_{n-1}\sqrt{6} $
$\frac{F(n)}{F(n-1)} = 5+2\sqrt{6} $
$ F(n) = (5+2\sqrt{6})F(n-1) $
$ F(n) = (5+2\sqrt{6})a_{n-1}+(5\sqrt{6}+12)b_{n-1} $
$ F(n) = a_n+b_n\sqrt{6} $
$ a_n = (5+2\sqrt{6})a_{n-1} $, $ b_n\sqrt{6} = (5\sqrt{6}+12)b_{n-1} $
$ a_n+b_n\sqrt{6} = (5a_{n-1}+12b_{n-1})+(2a_{n-1}+5b_{n-1})\sqrt{6} $
inline ll mod(ll a){return a%MOD;}
struct Matrix{
ll mt[22][22],r,c;
void init(int rr,int cc,bool flag=0){
r=rr;c=cc;
memset(mt,0,sizeof mt);
if(flag) rep(i,1,r) mt[i][i]=1;
}
Matrix operator * (const Matrix &rhs)const{
Matrix ans; ans.init(r,rhs.c);
rep(i,1,r){
rep(j,1,rhs.c){
int t=max(r,rhs.c);
rep(k,1,t){
ans.mt[i][j]+=mod(mt[i][k]*rhs.mt[k][j]);
ans.mt[i][j]=mod(ans.mt[i][j]);
}
}
}
return ans;
}
};
Matrix fpw(Matrix A,int n){
Matrix ans;ans.init(A.r,A.c,1);
while(n){
if(n&1) ans=ans*A;
n>>=1;
A=A*A;
}
return ans;
}
int bas[3][3]={
{0,0,0},
{0,5,12},
{0,2,5},
};
int bas2[]={0,5,2};
int main(){
Matrix A;A.init(2,2);
rep(i,1,2) rep(j,1,2) A.mt[i][j]=bas[i][j];
Matrix b; b.init(2,1);
rep(i,1,2) b.mt[i][1]=bas2[i];
int T=read();
while(T--){
int n=read();
Matrix res=fpw(A,n-1); res=res*b;
ll ans=mod(2*res.mt[1][1]-1+MOD);
println(ans);
}
return 0;
}