特殊矩陣的冪同樣滿足費馬小定理。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define int long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'
 
a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    
 
return x*f;
}
const int P=1125899839733759ll;
int T,n;
int ksc(int a,int b,int p)
{
    int t=a*b-(int)((long double)a*b/p+0.5)*p;
    return t<0?t+p:t;
}
int ksm(int a,int k,int p)
{
    int s=1;
    for (;k;k>>=1,a=ksc(a,a,p)) if (k&1) s=ksc(s,a,p);
    return s;
}
struct matrix
{
    int
 
 n,a[2][2];
    matrix operator *(const matrix&b) const
    {
        matrix c;c.n=n;memset(c.a,0,sizeof(c.a));
        for (int i=0;i<n;i++)
            for (int j=0;j<2;j++)
                for (int k=0;k<2;k++)
                c.a[i][j]=(c.a[i][j]+ksc(a[i][k],b.a[k][j],P))%P;
        return c;
    }
}f,a;
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5118.in","r",stdin);
    freopen("bzoj5118.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        n=ksm(2,read(),P-1);
        f.n=1;f.a[0][0]=0,f.a[0][1]=1;
        a.n=2;a.a[0][0]=0,a.a[0][1]=a.a[1][0]=a.a[1][1]=1;
        for (;n;n>>=1,a=a*a) if (n&1) f=f*a;
        cout<<f.a[0][0]<<endl;
    }
    return 0;
}