求解斐波那契數列

1：遞迴演算法：

long long Fib(int n) {
	if (n <= 1)	return n;
	return Fib(n - 1) + Fib(n - 2);
}

但這會導致大量重複的計算

2：記憶化搜尋/DP

long long num[100] = { 0 };

long long Fib(int n) {
	if (n <= 1)	return n;
	if (num[n])	return num[n];
	return num[n] = Fib(n - 1) + Fib(n - 2);
}

3：公式

long long Fib(int n) {
	double t = sqrt(5);
	return ( (pow((1 + t) / 2, n) - pow((1 - t) / 2, n)) ) / t;
}
 

4：矩陣快速冪

#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>

using namespace std;

const int N = 2;
long long tmp[N][N];
//矩陣乘法
void multi(long long a[][N], long long b[][N]){
	memset(tmp, 0, sizeof (tmp));
	for (int i = 0; i<2; i++)
		for (int j = 0; j<2; j++)
			for (int k = 0; k<2; k++)
				tmp[i][j] += a[i][k] * b[k][j];

	for (int i = 0; i<2; i++)
		for (int j = 0; j<2; j++)
			a[i][j] = tmp[i][j];
}

long long res[N][N];
//矩陣快速冪
void quickpow(long long a[][N], int n){
	memset(res, 0, sizeof res);
	for (int i = 0; i<N; i++) res[i][i] = 1;
	while (n){
		if (n & 1)
			multi(res, a);	//res=res*a;
		multi(a, a);	//a=a*a
		n >>= 1;
	}
}

//求解斐波那契數列
long long solve(int n) {
	long long m[2][2] = { 1, 1, 1, 0 };
	quickpow(m, n);

	return res[0][1];
}

int main() {
	cout << solve(10) << endl;
	
	return 0;
}