递归大总结之斐波那契

//利用循环的解法，效率高

#include<iostream>
using namespace std;

int facii(int n)
{
  if (n < 0)
  {
    return 0;
  }
  int a[] = { 0, 1 };
  int i = 0, x1 = 0, x2 = 1, x3 = 0;

  if (n<2)
    return a[n];

  for (i = 2; i <= n; i++)
  {
    x3 = x1 + x2;
    x1 = x2;
    x2 = x3;
  }
  return x3;
}

int main()
{
  int n, y;
  cin >> n;
  y = facii(n);
  cout << y << endl;
  system("pause");
}

递归解法

#include<iostream>
using namespace std;
int f(int n)
{
  if (n < 0)
  {
    return 0;
  }
  int arr[2] = { 0, 1 };
  if (n < 2)
  {  
    return arr[n];
  }
  return f(n - 1) + f(n - 2);
}
int main()
{
  int n;
  int res;
  cin >> n;
  res=f(n);
  cout << res << endl;
  system("pause");
}