【每日算法Day 80】所有人都会做的入门题，高级解法来了！-CFANZ编程社区

【每日算法Day 80】所有人都会做的入门题，高级解法来了！

题目还是昨天的题，昨天已经介绍了三种解法了，今天介绍一个最快的方法。

题目链接

LeetCode 面试题 08.01. 三步问题[1]

示例1

输入：
n = 3 
输出：
4
解释：
有四种走法

示例2

输入：
n = 5
输出：
13

代码

c++

typedef long long ll;
typedef vector<vector<ll>> vvl;
typedef vector<ll> vl;
const ll p = 1e9+7;

class Solution {
public:
    vvl mat_mul(vvl& A, vvl& B) {
        int a = A.size(), b = B.size(), c = B[0].size();
        vvl C(a, vl(c, 0));
        for (int i = 0; i < a; ++i) {
            for (int j = 0; j < c; ++j) {
                for (int k = 0; k < b; ++k) {
                    (C[i][j] += A[i][k] * B[k][j]) %= p;
                }
            }
        }
        return C;
    }

    vvl mat_pow(vvl& A, int n) {
        int m = A.size();
        vvl B(m, vl(m, 0));
        for (int i = 0; i < m; ++i) B[i][i] = 1;
        while (n > 0) {
            if (n&1) B = mat_mul(B, A);
            A = mat_mul(A, A);
            n >>= 1;
        }
        return B;
    }

    vvl mat_pow_rec(vvl& A, int n) {
        if (n == 1) return A;
        vvl B = mat_pow_rec(A, n/2);
        B = mat_mul(B, B);
        if (n&1) return mat_mul(A, B);
        return B;
    }

    int waysToStep(int n) {
        vl f = {1, 2, 4};
        if (n <= 3) return f[n-1];
        vvl A = {{0, 0, 1}, {1, 0, 1}, {0, 1, 1}};
        vvl B = mat_pow(A, n-3);
        ll res = 0;
        for (int i = 0; i < 3; ++i) {
            (res += f[i] * B[i][2]) %= p;
        }
        return res;
    }
};

python

import numpy as np

class Solution:
    def mat_pow(self, A, n):
        m = A.shape[0]
        B = np.eye(m, dtype=np.int64)
        while n > 0:
            if (n&1)!=0:
                B = np.mod(np.matmul(B, A), self.p).astype(np.int64)
            A = np.mod(np.matmul(A, A), self.p).astype(np.int64)
            n >>= 1
        return B;

    def waysToStep(self, n: int) -> int:
        self.p = int(1e9+7)
        f = [1, 2, 4]
        if n <= 3: return f[n-1]
        A = np.array([[0, 0, 1], [1, 0, 1], [0, 1, 1]], dtype=np.int64)
        B = self.mat_pow(A, n-3)
        res = 0
        for i in range(3):
            res += f[i] * B[i][2]
        return int(res%self.p)