使用元编程，用最简单的递归构造快速斐波那契数函数。

template <int T>
class metafib {
public:
    static const long long fib=metafib<T-1>::fib+metafib<T-2>::fib;
};

template <>
class metafib<1> {
public:
    static const int fib=1;
};

template <>
class metafib<2> {
public:
    static const int fib=2;
};

int main(){
    long a1=clock();
    cout<<metafib<200>::fib<<endl;
     long a2=clock();
    cout<<a2-a1;
}

long long fibcal(long long n){
    if(n<=2)
        return n;
    n-=2;
    long long a=1;
    long long b=2;
    while(n--){
        b=a+b;
        a=b-a;
    }
    return b;
};