Sep 13, 2023 · Sep 13, 2023 · Sep 13, 2023
diff --git a/solution/1100-1199/1137.N-th Tribonacci Number/README.md b/solution/1100-1199/1137.N-th Tribonacci Number/README.md
Original file line number	Diff line number	Diff line change
Expand Up		@@ -52,6 +52,32 @@ T_4 =わ 1 +たす 1 +たす 2 =わ 4

		时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为给定的整数。

	方法二:矩阵快速幂加速递推

	我们设 $Tib(n)$ 表示一个 1ドル \times 3$ 的矩阵 $\begin{bmatrix} T_n & T_{n - 1} & T_{n - 2} \end{bmatrix},ドル其中 $T_n,ドル $T_{n - 1}$ 和 $T_{n - 2}$ 分别表示第 $n$ 个、第 $n - 1$ 个和第 $n - 2$ 个泰波那契数。

	我们希望根据 $Tib(n-1) = \begin{bmatrix} T_{n - 1} & T_{n - 2} & T_{n - 3} \end{bmatrix}$ 推出 $Tib(n)$。也即是说,我们需要一个矩阵 $base,ドル使得 $Tib(n - 1) \times base = Tib(n),ドル即:

	$$
	\begin{bmatrix}
	T_{n - 1} & T_{n - 2} & T_{n - 3}
	\end{bmatrix} \times base = \begin{bmatrix} T_n & T_{n - 1} & T_{n - 2} \end{bmatrix}
	$$

	由于 $T_n = T_{n - 1} + T_{n - 2} + T_{n - 3},ドル所以矩阵 $base$ 为:

	$$
	\begin{bmatrix}
	1 & 1 & 0 \\
	1 & 0 & 1 \\
	1 & 0 & 0
	\end{bmatrix}
	$$

	我们定义初始矩阵 $res = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix},ドル那么 $T_n$ 等于 $res$ 乘以 $base^{n - 3}$ 的结果矩阵中所有元素之和。使用矩阵快速幂求解即可。

	时间复杂度 $O(\log n),ドル空间复杂度 $O(1)$。

		<!-- tabs:start -->

		### Python3
Expand All		@@ -67,6 +93,35 @@ class Solution:
		return a
		```

	```python
	class Solution:
	def tribonacci(self, n: int) -> int:
	def mul(a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
	m, n = len(a), len(b[0])
	c = [[0] * n for _ in range(m)]
	for i in range(m):
	for j in range(n):
	for k in range(len(a[0])):
	c[i][j] = c[i][j] + a[i][k] * b[k][j]
	return c

	def pow(a: List[List[int]], n: int) -> List[List[int]]:
	res = [[1, 1, 0]]
	while n:
	if n & 1:
	res = mul(res, a)
	n >>= 1
	a = mul(a, a)
	return res

	if n == 0:
	return 0
	if n < 3:
	return 1
	a = [[1, 1, 0], [1, 0, 1], [1, 0, 0]]
	return sum(pow(a, n - 3)[0])
	```

		### Java

		<!-- 这里可写当前语言的特殊实现逻辑 -->
Expand All		@@ -86,6 +141,51 @@ class Solution {
		}
		```

	```java
	class Solution {
	public int tribonacci(int n) {
	if (n == 0) {
	return 0;
	}
	if (n < 3) {
	return 1;
	}
	int[][] a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
	int[][] res = pow(a, n - 3);
	int ans = 0;
	for (int x : res[0]) {
	ans += x;
	}
	return ans;
	}

	private int[][] mul(int[][] a, int[][] b) {
	int m = a.length, n = b[0].length;
	int[][] c = new int[m][n];
	for (int i = 0; i < m; ++i) {
	for (int j = 0; j < n; ++j) {
	for (int k = 0; k < b.length; ++k) {
	c[i][j] += a[i][k] * b[k][j];
	}
	}
	}
	return c;
	}

	private int[][] pow(int[][] a, int n) {
	int[][] res = {{1, 1, 0}};
	while (n > 0) {
	if ((n & 1) == 1) {
	res = mul(res, a);
	}
	a = mul(a, a);
	n >>= 1;
	}
	return res;
	}
	}
	```

		### C++

		```cpp
Expand All		@@ -104,6 +204,50 @@ public:
		};
		```

	```cpp
	class Solution {
	public:
	int tribonacci(int n) {
	if (n == 0) {
	return 0;
	}
	if (n < 3) {
	return 1;
	}
	vector<vector<ll>> a = {{1, 1, 0}, {1, 0, 1}, {1, 0, 0}};
	vector<vector<ll>> res = pow(a, n - 3);
	return accumulate(res[0].begin(), res[0].end(), 0);
	}

	private:
	using ll = long long;
	vector<vector<ll>> mul(vector<vector<ll>>& a, vector<vector<ll>>& b) {
	int m = a.size(), n = b[0].size();
	vector<vector<ll>> c(m, vector<ll>(n));
	for (int i = 0; i < m; ++i) {
	for (int j = 0; j < n; ++j) {
	for (int k = 0; k < b.size(); ++k) {
	c[i][j] += a[i][k] * b[k][j];
	}
	}
	}
	return c;
	}

	vector<vector<ll>> pow(vector<vector<ll>>& a, int n) {
	vector<vector<ll>> res = {{1, 1, 0}};
	while (n) {
	if (n & 1) {
	res = mul(res, a);
	}
	a = mul(a, a);
	n >>= 1;
	}
	return res;
	}
	};
	```

		### Go

		```go
Expand All		@@ -116,6 +260,51 @@ func tribonacci(n int) int {
		}
		```

	```go
	func tribonacci(n int) (ans int) {
	if n == 0 {
	return 0
	}
	if n < 3 {
	return 1
	}
	a := [][]int{{1, 1, 0}, {1, 0, 1}, {1, 0, 0}}
	res := pow(a, n-3)
	for _, x := range res[0] {
	ans += x
	}
	return
	}

	func mul(a, b [][]int) [][]int {
	m, n := len(a), len(b[0])
	c := make([][]int, m)
	for i := range c {
	c[i] = make([]int, n)
	}
	for i := 0; i < m; i++ {
	for j := 0; j < n; j++ {
	for k := 0; k < len(b); k++ {
	c[i][j] += a[i][k] * b[k][j]
	}
	}
	}
	return c
	}

	func pow(a [][]int, n int) [][]int {
	res := [][]int{{1, 1, 0}}
	for n > 0 {
	if n&1 == 1 {
	res = mul(res, a)
	}
	a = mul(a, a)
	n >>= 1
	}
	return res
	}
	```

		### JavaScript

		```js
Expand All		@@ -137,6 +326,96 @@ var tribonacci = function (n) {
		};
		```

	```js
	/**
	* @param {number} n
	* @return {number}
	*/
	var tribonacci = function (n) {
	if (n === 0) {
	return 0;
	}
	if (n < 3) {
	return 1;
	}
	const a = [
	[1, 1, 0],
	[1, 0, 1],
	[1, 0, 0],
	];
	return pow(a, n - 3)[0].reduce((a, b) => a + b);
	};

	function mul(a, b) {
	const [m, n] = [a.length, b[0].length];
	const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
	for (let i = 0; i < m; ++i) {
	for (let j = 0; j < n; ++j) {
	for (let k = 0; k < b.length; ++k) {
	c[i][j] += a[i][k] * b[k][j];
	}
	}
	}
	return c;
	}

	function pow(a, n) {
	let res = [[1, 1, 0]];
	while (n) {
	if (n & 1) {
	res = mul(res, a);
	}
	a = mul(a, a);
	n >>= 1;
	}
	return res;
	}
	```

	### TypeScript

	```ts
	function tribonacci(n: number): number {
	if (n === 0) {
	return 0;
	}
	if (n < 3) {
	return 1;
	}
	const a = [
	[1, 1, 0],
	[1, 0, 1],
	[1, 0, 0],
	];
	return pow(a, n - 3)[0].reduce((a, b) => a + b);
	}

	function mul(a: number[][], b: number[][]): number[][] {
	const [m, n] = [a.length, b[0].length];
	const c = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
	for (let i = 0; i < m; ++i) {
	for (let j = 0; j < n; ++j) {
	for (let k = 0; k < b.length; ++k) {
	c[i][j] += a[i][k] * b[k][j];
	}
	}
	}
	return c;
	}

	function pow(a: number[][], n: number): number[][] {
	let res = [[1, 1, 0]];
	while (n) {
	if (n & 1) {
	res = mul(res, a);
	}
	a = mul(a, a);
	n >>= 1;
	}
	return res;
	}
	```

		### PHP

		```php
Expand Down