超级楼梯-斐波那契数列的运用

 1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int n, i, j;
 6     int a[40];
 7     a[2] = 1;
 8     a[3] = 2;
 9     for (i = 4; i <= 40; ++i)
10         a[i] = a[i - 1] + a[i - 2];
11     cin >> n;
12     for (i = 0; i<n; ++i)
13     {
14         cin >> j;
15         cout << a[j] << endl;
16     }
17 }