斐波那契數列的五種解法

定義 a₀=1，a₁=1，a_n = a_n−1 + a_n−2，求 a_n是多少。

爲了避免考慮整數溢出問題，我們求 a_n % p 的值，p=10⁹+7。

遞歸

const int mod = 1000000007;
void f(int n)
{
	if (n <= 1) return 1;
	return (f(n - 1) + f(n - 2)) % mod;
}

記憶化搜索

const int N = 100005;
const int mod = 1000000007;
int a[N];
int f(int n)
{
	if (a[n]) return a[n];
	if (n <= 1) return 1;
	a[n] = f(n - 1) + f(n - 2);
	a[n] %= mod;
	return a[n];
}

遞推

const int N = 100005, mod = 1000000007;
int f(int n)
{
    a[0] = a[1] = 1;
    for (int i = 2; i <= n; i ++ )
    {
        a[i] = a[i - 1] + a[i - 2];
        a[i] %= mod;
    }
    return a[n];
}

遞歸+滾動變量

const int mod = 1000000007;
void f(int n)
{
	int a, b;
	a = b = 1;
    for (int i = 2; i <= n; i++)
    {
    	int c = (a + b) % mod;
    	a = b;
    	b = c;
	}
}

矩陣運算 + 快速冪

typedef long long ll;
const ll mod = 100000007;
struct mat{
	ll a[2][2];
};
mat poww(mat A, ll n)
{
	mat res;
	res.a[0][0] = 1, res.a[0][1] = 0, res.a[1][0] = 0, res.m[1][1] = 1;
	while (n)
	{
		if (n & 1)
		{
			res = mul(res, A);
		}
		A = mul(A, A);
		n >>= 1;
	}
	return res;
}
mat mul(mat x, mat y)
{
	mat res;
	memset(res.a, 0, sizeof(res.a));
	for (int i = 0; i < 2; i++)
	{
		for (int j = 0; j < 2; j++)
		{
			for (int k = 0; k < 2; k++)
			{
				res.a[i][j] += x.a[i][k] * y.a[k][j] % mod;
			}
		}
	}
	return res;
}