600. 不含连续1的非负整数

摘要
Title: 600. 不含连续1的非负整数
Categories: 数位dp

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600. 不含连续1的非负整数

题意

给定一个正整数 n ,请你统计在 [0, n] 范围的非负整数中,有多少个整数的二进制表示中不存在 连续的 1 。

思路

套用模板,但由于这里包括了0,所以0要特判一下
具体思路就是将n转换为二进制进行数位dp,last改为上一位是什么

代码

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'''
Author: NEFU AB-IN
Date: 2024-08-13 14:15:44
FilePath: \LeetCode\600\600.py
LastEditTime: 2024-08-13 15:08:08
'''
# 3.8.19 import
import random
from collections import Counter, defaultdict, deque
from datetime import datetime, timedelta
from functools import lru_cache, reduce
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from itertools import combinations, compress, permutations, starmap, tee
from math import ceil, comb, fabs, floor, gcd, hypot, log, perm, sqrt
from string import ascii_lowercase, ascii_uppercase
from sys import exit, setrecursionlimit, stdin
from typing import Any, Callable, Dict, List, Optional, Tuple, TypeVar, Union

# Constants
TYPE = TypeVar('TYPE')
N = int(2e5 + 10)
M = int(20)
INF = int(1e12)
OFFSET = int(100)
MOD = int(1e9 + 7)

# Set recursion limit
setrecursionlimit(int(2e9))


class Arr:
array = staticmethod(lambda x=0, size=N: [x() if callable(x) else x for _ in range(size)])
array2d = staticmethod(lambda x=0, rows=N, cols=M: [Arr.array(x, cols) for _ in range(rows)])
graph = staticmethod(lambda size=N: [[] for _ in range(size)])


class Math:
max = staticmethod(lambda a, b: a if a > b else b)
min = staticmethod(lambda a, b: a if a < b else b)


class IO:
input = staticmethod(lambda: stdin.readline().rstrip("\r\n"))
read = staticmethod(lambda: map(int, IO.input().split()))
read_list = staticmethod(lambda: list(IO.read()))


class Std:
pass

# ————————————————————— Division line ——————————————————————


class Solution:
def findIntegers(self, n: int) -> int:

s = bin(n)[2:]

@lru_cache(None)
def dfs(i, last, is_limit, is_num):
if i == n.bit_length():
return int(is_num)
res = 0
if not is_num:
res = dfs(i + 1, 0, False, False)

up = int(s[i]) if is_limit else 1
down = 0 if is_num else 1
for d in range(down, up + 1):
if d == 1 and last == 1:
continue
res += dfs(i + 1, d, is_limit and up == d, True)
return res

return dfs(0, 0, True, False) + 1


# Solution().findIntegers(10)

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