【100题】斐波那契数列(2)

//斐波那契数列#include <iostream>using namespace std;//方法2：递推，为了避免递归中的重复计算，我们可以从小的往大的计算。这样就没有重复的计算了long long Fibonacci(unsigned int n){int a[2] = {0,1};if(n < 2){return a[n];}long long b = 1;long long c = 0;long long z = 0;for(int i=2; i<=n; ++i){z = b + c;c = b;b = z;}return z;}void main(){cout << Fibonacci(7) <<endl;}