Recurrence Relations

FJNU.1189

Description
Recurrence relations are where a function is defined in terms of itself and maybe other functions as well. in this problem, we have two functions of interest:
F (N) = F (N - 1) + G (N - 1) ; F (1) = 1
G (N) = G (N - 1) + G (N - 3) ; G (1) = 1, G (2) = 0, G (3) = 1
For a given value of N, compute F (N).

Input
Each line of input will have exactly one integer, 57 > = N > 0.

Output
For each line of input, output F(N).

Sample Input
1
4
57

Sample Output
1
3
2035586497

My Program

#include<iostream>
using namespace std;

int main()
...{
    long a[58]=...{0,1,2,2,3,5,7,10,15,22,32,47,69,101,148,217,318,466,683,1001,1467,2150,3151,4618,6768,9919,14537,21305,31224,45761,67066,98290,144051,211117,309407,453458,664575,973982,1427440,2092015,3065997,4493437,6585452,9651449,14144886,20730338,30381787,44526673,65257011,95638798,140165471,205422482,301061280,441226751,646649233,947710513,1388937264,2035586497};
    int n;
    while(cin>>n)
        cout<<a[n]<<endl;
    return 0;
}

附带递归计算程序

#include<iostream>
using namespace std;

long G(int n)
...{
    if(n==1)
        return 1;
    else
        if(n==2)
            return 0;
        else
            if(n==3)
                return 1;
            else
                return G(n-1)+G(n-3);
}

long F(int n)
...{
    if(n==1)
        return 1;
    else
        return F(n-1)+G(n-1);
}

int main()
...{
    long n;
    int k;
    while(cin>>k)
    ...{
        n=F(k);
        cout<<n<<endl;
    }
    return 0;
}

YOYO's Note: 
打表。
直接递归算肯定会TLE，反正它都给了0<N<=57，不打白不打……
打到50左右的时候每个都要等好久才出来……
幸好位数不是很多，直接long解决问题。