KMP+矩阵快速幂模版

题目链接

https://leetcode.cn/problems/string-transformation/solutions/2435348/kmp-ju-zhen-kuai-su-mi-you-hua-dp-by-end-vypf/

模版代码

class Solution:
    def numberOfWays(self, s, t, k):
        n = len(s)
        c = self.kmp_search(s + s[:-1], t)
        #print(c)
        m = [
            [c - 1, c],
            [n - c, n - 1 - c]
        ]
        m = self.pow(m, k)
        return m[0][s != t]

    # KMP 模板
    def calc_max_match(self, s: str) -> List[int]:
        match = [0] * len(s)
        c = 0
        for i in range(1, len(s)):
            v = s[i]
            while c and s[c] != v:
                c = match[c - 1]
            if s[c] == v:
                c += 1
            match[i] = c
        return match

    # KMP 模板
    # 返回 text 中出现了多少次 pattern（允许 pattern 重叠）
    def kmp_search(self, text: str, pattern: str) -> int:
        match = self.calc_max_match(pattern)
        match_cnt = c = 0
        for i, v in enumerate(text):
            v = text[i]
            while c and pattern[c] != v:
                c = match[c - 1]
            if pattern[c] == v:
                c += 1
            if c == len(pattern):
                match_cnt += 1
                c = match[c - 1]
        return match_cnt

    # 矩阵乘法
    def multiply(self, a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
        c = [[0, 0], [0, 0]]
        for i in range(2):
            for j in range(2):
                c[i][j] = (a[i][0] * b[0][j] + a[i][1] * b[1][j]) % (10 ** 9 + 7)
        return c

    # 矩阵快速幂
    def pow(self, a: List[List[int]], n: int) -> List[List[int]]:
        res = [[1, 0], [0, 1]]
        while n:
            if n % 2:
                res = self.multiply(res, a)
            a = self.multiply(a, a)
            n //= 2
        return res

kmp作用

快速找出text中匹配pattern的次数

矩阵快速幂作用

找出dp递推式，需要求矩阵的k次方，k很大，使用矩阵快速幂