#include<iostream>
#include<vector>
using namespace std;/**************************************
*        
*        斐波那契問題的三種解法
*
***************************************/
//方法1：遞歸調用函數,時間復雜度 = 2^N
int Fibonacci_fun1(int n)
{if (n < 1)return 0;else if (n == 1 || n == 2)return 1;elsereturn Fibonacci_fun1(n - 1) + Fibonacci_fun1(n - 2);
}
//方法2：非遞歸方法，時間復雜度 = N
int Fibonacci_fun2(int n)
{if (n < 1)return 0;else if (n == 1 || n == 2)return 1;else{int res = 1;int pre = 1;int temp = 0;for (int i = 3; i <= n; ++i){temp = res;res = pre;pre = pre + temp;}return pre;}
}
//方法3：矩陣乘法，時間復雜度 = logN
vector<vector<int>> Mat_multip(vector<vector<int>> m, vector<vector<int>> n)
{vector<vector<int>> res(m.size());for (int i = 0; i < m.size(); ++i)res[i].reserve(2);int temp = 0;for (int i = 0; i < m.size(); ++i){for (int j = 0; j < n.at(0).size(); ++j){temp = 0;for (int k = 0; k < m.at(0).size(); ++k){temp += m.at(i).at(k)*n.at(k).at(j);}res.at(i).push_back(temp);}}return res;
}vector<vector<int>> MatPower( vector<vector<int>> m, int n)
{vector<vector<int>> res = { {1,0},{0,1} };vector<vector<int>> temp = m;for (; n != 0; n >>= 1){if ((n & 0x1) == 1)res = Mat_multip(res, temp);temp = Mat_multip(temp, temp);}return res;
}int Fibonacci_fun3(int n)
{if (n < 1)return 0;else if (n == 1 || n == 2)return 1;else{vector<vector<int>> base = { { 1,1 },{ 1,0 } };vector<vector<int>> res;res = MatPower(base, n - 2);return res[0][0] + res[1][0];}
}int main()
{cout << "Fibonacci_fun2: " << Fibonacci_fun3(40) << endl;cout << "Fibonacci_fun2: " << Fibonacci_fun2(40) << endl;cout << "Fibonacci_fun1: " << Fibonacci_fun1(40) << endl;return 0;
}

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