【C++】斐波那契数列
#include<iostream>using namespace std;
const int N = 100010;
int fbnq(int n)
{
if (n == 1 || n == 2) return 1;
if(n > 2)
return (fbnq(n - 1) + fbnq(n - 2));
}
int main ()
{
int n;
cin >> n;
cout << fbnq(n) << endl;
}
原有2只小兔,小兔长成大兔一月,有2只大兔就会生只小兔。兔兔只数:2,2,4,6,10,16,26,42......
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