bits.go

package copypasta

import (
	. "fmt"
	"math"
	"math/bits"
	"slices"
)

/*
从集合论到位运算，常见位运算技巧分类总结！
https://leetcode.cn/circle/discuss/CaOJ45/

有关二进制枚举、枚举子集的子集、枚举大小固定集合等写法，见 search.go

运算符优先级 https://golang.org/ref/spec#Operators
Precedence    Operator
    5         *  /  %  <<  >>  &  &^
    4         +  -  |  ^
    3         ==  !=  <  <=  >  >=
    2         &&
    1         ||

标准库 "math/bits" 包含了位运算常用的函数，如二进制中 1 的个数、二进制表示的长度等
注意：bits.Len(0) 返回的是 0 而不是 1
     bits.Len(x) 相当于 int(Log2(x)+eps)+1  x>0
     或者说 2^(Len(x)-1) <= x < 2^Len(x)    x>0

### 基础题
- [1486. 数组异或操作](https://leetcode.cn/problems/xor-operation-in-an-array/) 1181
- [2595. 奇偶位数](https://leetcode.cn/problems/number-of-even-and-odd-bits/) 1207
- [231. 2 的幂](https://leetcode.cn/problems/power-of-two/)
- [342. 4 的幂](https://leetcode.cn/problems/power-of-four/)
- [476. 数字的补数](https://leetcode.cn/problems/number-complement/) 1235
- [191. 位 1 的个数](https://leetcode.cn/problems/number-of-1-bits/)
- [338. 比特位计数](https://leetcode.cn/problems/counting-bits/) 也可以 DP
- [1356. 根据数字二进制下 1 的数目排序](https://leetcode.cn/problems/sort-integers-by-the-number-of-1-bits/) 1258
- [461. 汉明距离](https://leetcode.cn/problems/hamming-distance/)
- [2220. 转换数字的最少位翻转次数](https://leetcode.cn/problems/minimum-bit-flips-to-convert-number/) 1282
- [868. 二进制间距](https://leetcode.cn/problems/binary-gap/) 1307
- [2917. 找出数组中的 K-or 值](https://leetcode.cn/problems/find-the-k-or-of-an-array/) 1389
- [693. 交替位二进制数](https://leetcode.cn/problems/binary-number-with-alternating-bits/)

### 与或（AND/OR）的性质
TIPS: & 和 | 在区间求和上具有单调性；^ 的区间求和见 strings.go 中的 trie.maxXor
** 代码和题目见下面的 bitOpTrick 和 bitOpTrickCnt
- [2980. 检查按位或是否存在尾随零](https://leetcode.cn/problems/check-if-bitwise-or-has-trailing-zeros/) 1234
- [1318. 或运算的最小翻转次数](https://leetcode.cn/problems/minimum-flips-to-make-a-or-b-equal-to-c/) 1383
- [2419. 按位与最大的最长子数组](https://leetcode.cn/problems/longest-subarray-with-maximum-bitwise-and/) 1496
- [2871. 将数组分割成最多数目的子数组](https://leetcode.cn/problems/split-array-into-maximum-number-of-subarrays/) 1750
- [2401. 最长优雅子数组](https://leetcode.cn/problems/longest-nice-subarray/) 1750
- [2680. 最大或值](https://leetcode.cn/problems/maximum-or/) 1912 可以做到 $\mathcal{O}(1)$ 额外空间
- [2411. 按位或最大的最小子数组长度](https://leetcode.cn/problems/smallest-subarrays-with-maximum-bitwise-or/) 1938
- [898. 子数组按位或操作](https://leetcode.cn/problems/bitwise-ors-of-subarrays/) 2133
- [1521. 找到最接近目标值的函数值](https://leetcode.cn/problems/find-a-value-of-a-mysterious-function-closest-to-target/) 2384
两数 OR 的最小值：只需要知道区间内最小的 bits.Len(U) + 1 个数 https://codeforces.com/problemset/problem/1665/E

### 异或（XOR）的性质
- [1720. 解码异或后的数组](https://leetcode.cn/problems/decode-xored-array/) 1284
- [2433. 找出前缀异或的原始数组](https://leetcode.cn/problems/find-the-original-array-of-prefix-xor/) 1367
- [1310. 子数组异或查询](https://leetcode.cn/problems/xor-queries-of-a-subarray/) 1460
- [2683. 相邻值的按位异或](https://leetcode.cn/problems/neighboring-bitwise-xor/) 1518
- [1829. 每个查询的最大异或值](https://leetcode.cn/problems/maximum-xor-for-each-query/) 1523
- [2997. 使数组异或和等于 K 的最少操作次数](https://leetcode.cn/problems/minimum-number-of-operations-to-make-array-xor-equal-to-k/) 1525
- [1442. 形成两个异或相等数组的三元组数目](https://leetcode.cn/problems/count-triplets-that-can-form-two-arrays-of-equal-xor/) 1525
- [2429. 最小异或](https://leetcode.cn/problems/minimize-xor/) 1532
- [2527. 查询数组异或美丽值](https://leetcode.cn/problems/find-xor-beauty-of-array/) 1550
- [2317. 操作后的最大异或和](https://leetcode.cn/problems/maximum-xor-after-operations/) 1679
- [2588. 统计美丽子数组数目](https://leetcode.cn/problems/count-the-number-of-beautiful-subarrays/) 1697
- [2564. 子字符串异或查询](https://leetcode.cn/problems/substring-xor-queries/) 1959
- [1734. 解码异或后的排列](https://leetcode.cn/problems/decode-xored-permutation/) 2024
- [2857. 统计距离为 k 的点对](https://leetcode.cn/problems/count-pairs-of-points-with-distance-k/) 2082
https://codeforces.com/problemset/problem/1895/D
https://codeforces.com/problemset/problem/1790/E

### 利用 lowbit
https://codeforces.com/problemset/problem/1689/E
交互 https://codeforces.com/problemset/problem/1780/D

### 拆位 / 贡献法（部分题目排序很有用）
- [477. 汉明距离总和](https://leetcode.cn/problems/total-hamming-distance/)
- [1863. 找出所有子集的异或总和再求和](https://leetcode.cn/problems/sum-of-all-subset-xor-totals/) 可以做到 $\mathcal{O}(n)$ 时间
- [2425. 所有数对的异或和](https://leetcode.cn/problems/bitwise-xor-of-all-pairings/) 1622 可以做到 $\mathcal{O}(n+m)$ 时间
- [2275. 按位与结果大于零的最长组合](https://leetcode.cn/problems/largest-combination-with-bitwise-and-greater-than-zero/) 1642
- [1835. 所有数对按位与结果的异或和](https://leetcode.cn/problems/find-xor-sum-of-all-pairs-bitwise-and/) 1825 也有恒等式做法
- [2505. 所有子序列和的按位或](https://leetcode.cn/problems/bitwise-or-of-all-subsequence-sums/)（会员题）
《灵茶八题》完整题目列表 & 题解
https://www.luogu.com.cn/blog/endlesscheng/post-ling-cha-ba-ti-ti-mu-lie-biao
+ 表示元素和
^ 表示异或和
所有子数组的 + 的 + https://www.luogu.com.cn/problem/U360300
所有子数组的 ^ 的 ^ https://www.luogu.com.cn/problem/U360487
所有子数组的 ^ 的 + https://www.luogu.com.cn/problem/U360489
所有子数组的 + 的 ^ https://www.luogu.com.cn/problem/U360500
所有子序列的 + 的 + https://www.luogu.com.cn/problem/U360640
所有子序列的 ^ 的 ^ https://www.luogu.com.cn/problem/U360641
所有子序列的 ^ 的 + https://www.luogu.com.cn/problem/U360642 LC1863 https://leetcode.cn/problems/sum-of-all-subset-xor-totals/
所有子序列的 + 的 ^ https://www.luogu.com.cn/problem/U360643
所有子数组的 ^2 的 + 的 + https://www.nowcoder.com/feed/main/detail/857f180290cd402ea2461b85e94b3db9
- 这里 ^2 表示子数组中任意两个数的异或
所有子序列的 + 的 | LC2505 https://leetcode.cn/problems/bitwise-or-of-all-subsequence-sums/
https://www.lanqiao.cn/problems/10010/learning/?contest_id=157

https://codeforces.com/problemset/problem/1777/F
https://codeforces.com/problemset/problem/981/D
https://codeforces.com/problemset/problem/1895/D 1900
https://atcoder.jp/contests/abc281/tasks/abc281_f
https://atcoder.jp/contests/arc127/tasks/arc127_d
考虑贡献 https://codeforces.com/problemset/problem/1362/C 1400

### 试填法
- [3007. 价值和小于等于 K 的最大数字](https://leetcode.cn/problems/maximum-number-that-sum-of-the-prices-is-less-than-or-equal-to-k/) 2258
- [421. 数组中两个数的最大异或值](https://leetcode.cn/problems/maximum-xor-of-two-numbers-in-an-array/)，[试填法题解](https://leetcode.cn/problems/maximum-xor-of-two-numbers-in-an-array/solution/tu-jie-jian-ji-gao-xiao-yi-tu-miao-dong-1427d/)
- [2935. 找出强数对的最大异或值 II](https://leetcode.cn/problems/maximum-strong-pair-xor-ii/) 2349
- [3022. 给定操作次数内使剩余元素的或值最小](https://leetcode.cn/problems/minimize-or-of-remaining-elements-using-operations/) 2918
https://codeforces.com/contest/1918/problem/C 1400
加法拆位（进位拆位）：涉及到加法进位的题目，可以按照 mod 2^k 拆位
https://atcoder.jp/contests/abc091/tasks/arc092_b
所有 a[i]+a[j] 的异或和 https://codeforces.com/problemset/problem/1322/B 2100
变形：减法拆位（借位拆位）https://www.luogu.com.cn/problem/P3760
拆位再合并相同位 https://codeforces.com/problemset/problem/1874/B

### 恒等式
结合律：(a&b)^(a&c) = a&(b^c)    其他符号类似
- [1835. 所有数对按位与结果的异或和](https://leetcode.cn/problems/find-xor-sum-of-all-pairs-bitwise-and/) 1825
集合论公式的二进制等价形式：
popcount(a&b) + popcount(a|b) = popcount(a) + popcount(b)
- [2354. 优质数对的数目](https://leetcode.cn/problems/number-of-excellent-pairs/) 2076
https://oeis.org/A006234 (n+2) * 3^(n-2)   [0,2^n) 内任意两数 popcount(x) + popcount(y) - popcount(x+y) = 1 的数对个数
- https://codeforces.com/problemset/problem/1761/D
进位与分类讨论 https://codeforces.com/problemset/problem/1761/D https://www.luogu.com.cn/blog/linyihdfj/solution-cf1761d https://www.cnblogs.com/linyihdfj/p/16893607.html
a|b = (a^b) + (a&b)
a&b = (a|b) - (a^b)
a^b = (a|b) - (a&b)
a+b = (a|b) + (a&b)
    = (a&b)*2 + (a^b)
    = (a|b)*2 - (a^b)
(a^b) & (a&b) = 0 恒成立
https://codeforces.com/problemset/problem/76/D
https://codeforces.com/problemset/problem/1325/D
https://codeforces.com/problemset/problem/1368/D
https://codeforces.com/problemset/problem/1790/E
https://atcoder.jp/contests/abc050/tasks/arc066_b
a|b = (^a)&b + a
+ 与 ^ https://codeforces.com/problemset/problem/1732/C2
进位的本质 https://atcoder.jp/contests/arc158/tasks/arc158_c
max(a,b) = (a + b + abs(a-b)) / 2
min(a,b) = (a + b - abs(a-b)) / 2

### 思维题（贪心、脑筋急转弯等）
- [2546. 执行逐位运算使字符串相等](https://leetcode.cn/problems/apply-bitwise-operations-to-make-strings-equal/) 1605
- [1558. 得到目标数组的最少函数调用次数](https://leetcode.cn/problems/minimum-numbers-of-function-calls-to-make-target-array/) 1637
- [2571. 将整数减少到零需要的最少操作数](https://leetcode.cn/problems/minimum-operations-to-reduce-an-integer-to-0/) 1649 巧妙结论
- [2568. 最小无法得到的或值](https://leetcode.cn/problems/minimum-impossible-or/) 1754
- [2939. 最大异或乘积](https://leetcode.cn/problems/maximum-xor-product/) 2128
- [2749. 得到整数零需要执行的最少操作数](https://leetcode.cn/problems/minimum-operations-to-make-the-integer-zero/) 2132
- [2835. 使子序列的和等于目标的最少操作次数](https://leetcode.cn/problems/minimum-operations-to-form-subsequence-with-target-sum/) 2207
    - 相似题目 https://codeforces.com/problemset/problem/1918/C
- [2897. 对数组执行操作使平方和最大](https://leetcode.cn/problems/apply-operations-on-array-to-maximize-sum-of-squares/) 2301
- [810. 黑板异或游戏](https://leetcode.cn/problems/chalkboard-xor-game/) 2341
https://codeforces.com/problemset/problem/309/C 1900

### 其它
- [136. 只出现一次的数字](https://leetcode.cn/problems/single-number/)
- [287. 寻找重复数](https://leetcode.cn/problems/find-the-duplicate-number/)
- [260. 只出现一次的数字 III](https://leetcode.cn/problems/single-number-iii/)
- [137. 只出现一次的数字 II](https://leetcode.cn/problems/single-number-ii/)
- [645. 错误的集合](https://leetcode.cn/problems/set-mismatch/)
- [190. 颠倒二进制位](https://leetcode.cn/problems/reverse-bits/)
- [371. 两整数之和](https://leetcode.cn/problems/sum-of-two-integers/)
- [201. 数字范围按位与](https://leetcode.cn/problems/bitwise-and-of-numbers-range/)
- [2154. 将找到的值乘以 2](https://leetcode.cn/problems/keep-multiplying-found-values-by-two/) 可以做到 $\mathcal{O}(n)$ 时间
- [2044. 统计按位或能得到最大值的子集数目](https://leetcode.cn/problems/count-number-of-maximum-bitwise-or-subsets/) 1568
- [2438. 二的幂数组中查询范围内的乘积](https://leetcode.cn/problems/range-product-queries-of-powers/) 1610
- [1680. 连接连续二进制数字](https://leetcode.cn/problems/concatenation-of-consecutive-binary-numbers/) 1630
- [982. 按位与为零的三元组](https://leetcode.cn/problems/triples-with-bitwise-and-equal-to-zero/) 2085
- [1611. 使整数变为 0 的最少操作次数](https://leetcode.cn/problems/minimum-one-bit-operations-to-make-integers-zero/) 2345

### 位运算与字符串
LC3019 https://leetcode.cn/problems/number-of-changing-keys/
https://codeforces.com/contest/691/problem/B
https://codeforces.com/contest/1907/problem/B

构造 2^n-1，即 n 个 1 的另一种方法: ^(-1<<n)

https://oeis.org/A060142 每一段连续 0 的长度均为偶数的数：如 100110000100
Ordered set S defined by these rules: 0 is in S and if x is in S then 2x+1 and 4x are in S
0, 1, 3, 4, 7, 9, 12, 15, 16, 19, 25, 28, 31, 33, 36, 39, 48, 51, 57, 60, 63, 64, 67, 73, 76, 79, 97, 100
https://oeis.org/A086747 Baum-Sweet sequence
相关题目：蒙德里安的梦想 https://www.acwing.com/problem/content/293/

https://oeis.org/A048004 最长连续 1 为 k 的长为 n 的二进制串的个数
相关题目：https://codeforces.com/problemset/problem/1027/E

https://oeis.org/A047778 Concatenation of first n numbers in binary, converted to base 10
相关题目 LC1680 https://leetcode.cn/problems/concatenation-of-consecutive-binary-numbers/
钱珀瑙恩数 Champernowne constant https://en.wikipedia.org/wiki/Champernowne_constant

https://oeis.org/A072339
Any number n can be written (in two ways, one with m even and one with m odd) in the form n = 2^k_1 - 2^k_2 + 2^k_3 - ... + 2^k_m
where the signs alternate and k_1 > k_2 > k_3 > ... >k_m >= 0; sequence gives minimal value of m
https://codeforces.com/problemset/problem/1617/E

异或和相关
https://atcoder.jp/contests/abc171/tasks/abc171_e
https://oeis.org/A003987 异或矩阵
https://oeis.org/A003815 0^1^2^3^...^n: a(0)=0, a(4n+1)=1, a(4n+2)=4n+3, a(4n+3)=0, a(4n+4)=4n+4
- https://codeforces.com/problemset/problem/1493/E
- https://codeforces.com/problemset/problem/460/D
- https://atcoder.jp/contests/abc121/tasks/abc121_d
- https://atcoder.jp/contests/arc133/tasks/arc133_d
https://oeis.org/A145768 异或和 i*i
https://oeis.org/A126084 异或和 质数
https://oeis.org/A018252 异或和 合数?
https://oeis.org/A072594 异或和 质因数分解 是积性函数 a(p^k)=p*(k&1)
- https://oeis.org/A072595 满足 A072594(n)=0 的数
https://oeis.org/A178910 异或和 因子
- https://oeis.org/A178911 满足 A178910(n)=n 的数 Perfex number

异或与 mex
[1800·hot10] https://codeforces.com/problemset/problem/1554/C

异或与 <
a < b，无法通过两边异或同一个数来做式子变形
此时可以枚举高 k 个比特位是相等的，而第 k+1 个比特位 a 中是 0，b 中是 1
人为地创造出「相等」这个条件
https://codeforces.com/problemset/problem/1720/D2

路径点权异或 https://codeforces.com/problemset/problem/1709/E

https://oeis.org/A038712 a(n) = n^(n-1) = 1, 3, 1, 7, 1, 3, 1, 15, 1, ...
https://oeis.org/A080277 A038712 的前缀和  =>  a(n) = n + 2*a(n/2)

https://oeis.org/A055944 a(n) = n + (reversal of base-2 digits of n) (written in base 10)

二进制长度
https://oeis.org/A029837 Binary order of n: log_2(n) rounded up to next integer
https://oeis.org/A001855 A029837 的前缀和
https://oeis.org/A070939 a(0)=1, a(n)=bits.Len(n)
https://oeis.org/A083652 A070939 的前缀和

OnesCount 相当于二进制的 digsum
https://oeis.org/A000120 wt(n) = OnesCount(n)
https://oeis.org/A000788 前缀和 a(0) = 0, a(2n) = a(n)+a(n-1)+n, a(2n+1) = 2a(n)+n+1
https://oeis.org/A121853 前缀积 https://www.luogu.com.cn/problem/P4317
https://oeis.org/A092391 n+OnesCount(n)
	https://oeis.org/A010061 二进制自我数/哥伦比亚数（A092391 的补集）
https://oeis.org/A011371 n-OnesCount(n) Also highest power of 2 dividing n!
							a(n) = floor(n/2) + a(floor(n/2))
                         这同时是前 n 个数的质因子分解的 2 的幂次之和
https://oeis.org/A027868 Number of trailing zeros in n!; highest power of 5 dividing n!
                            a(n) = (n-A053824(n))/4, 其中 A053824(n) = Sum of digits of (n written in base 5)
推广至任意数：n! 的质因子分解中，p 的幂次为 (n-digsum_p(n))/(p-1)，其中 digsum_p(n) 表示 n 的 p 进制的数位和
https://oeis.org/A245788 n*OnesCount(n)
https://oeis.org/A049445 OnesCount(n)|n
	-  n/OnesCount(n)
https://oeis.org/A199238 n%OnesCount(n)
https://oeis.org/A010062 a(0)=1, a(n+1)=a(n)+OnesCount(a(n))
	https://oeis.org/A096303 从 n 出发不断执行 n+=OnesCount(n)，直到 n 在 A010062 中，所需要的迭代次数
	Number of iterations of n -> n + (number of 1's in binary representation of n) needed for the trajectory of n to join the trajectory of A010062
		https://oeis.org/A229743 Positions of records
		https://oeis.org/A229744 Values of records
	相关题目 https://www.luogu.com.cn/problem/P5891 https://class.luogu.com.cn/classroom/lgr66
https://oeis.org/A180094 Number of steps to reach 0 or 1, starting with n and applying the map k -> (number of 1's in binary expansion of k) repeatedly

https://oeis.org/A023416 Number of 0's in binary expansion of n
							a(n) = a(n/2) + 1 - n&1
https://oeis.org/A059015 A023416 的前缀和

十进制 digsum
一点点数学 https://codeforces.com/problemset/problem/817/C
https://oeis.org/A007953 digsum(n)
https://oeis.org/A062028 n+digsum(n)    质数 https://oeis.org/A047791    合数 https://oeis.org/A107743
	https://oeis.org/A003052 自我数/哥伦比亚数 Self number / Colombian number
	https://en.wikipedia.org/wiki/Self_number
	1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, 97, 108, ...
		https://oeis.org/A006378 自我质数 Self primes
https://oeis.org/A066568 n-digsum(n)
https://oeis.org/A057147 n*digsum(n)
https://oeis.org/A005349 digsum(n)|n   Niven (or Harshad) numbers
	https://oeis.org/A065877 digsum(n)∤n   Non-Niven (or non-Harshad) numbers
	https://oeis.org/A001101 Moran numbers: n such that (n / digsum(n)) is prime
https://oeis.org/A016052 a(1)=3, a(n+1)=a(n)+digsum(a(n))
https://oeis.org/A051885 Smallest number whose digsum = n
							(n%9+1) * int(math.Pow10(n/9)) - 1
							相关题目 https://codeforces.com/contest/1373/problem/E
https://oeis.org/A077196 Smallest possible sum of the digits of a multiple of n https://oeis.org/A077194 https://oeis.org/A077195
							相关题目（0-1 最短路）https://atcoder.jp/contests/arc084/tasks/arc084_b
https://oeis.org/A118137 digsum(n)+digsum(n+1)
https://oeis.org/A003132 Sum of squares of digits of n
	https://oeis.org/A003621 Number of iterations until n reaches 1 or 4 under x goes to sum of squares of digits map
https://oeis.org/A055012 Sum of cubes of digits of n
https://oeis.org/A055013 Sum of 4th powers of digits of n
https://oeis.org/A055014 Sum of 5th powers of digits of n
https://oeis.org/A055015 Sum of 6th powers of digits of n
	相关题目 https://www.luogu.com.cn/problem/P1660
https://oeis.org/A031286 Additive persistence: number of summations of digits needed to obtain a single digit (the additive digital root)
https://oeis.org/A031346 Multiplicative persistence: number of iterations of "multiply digits" needed to reach a number < 10

https://oeis.org/A014837 Sum of all the digits of n in every base from 2 to n-1
https://oeis.org/A043306 Sum of all the digits of n in every base from 2 to n

回文数
https://oeis.org/A002113 十进制回文数
	https://oeis.org/A043269 digsum(A002113(n))
	https://oeis.org/A070199 Number of palindromes of length <= n
https://oeis.org/A002779 回文平方数
	https://oeis.org/A002778 Numbers whose square is a palindrome
https://oeis.org/A002781 回文立方数
	https://oeis.org/A002780 Numbers whose cube is a palindrome
https://oeis.org/A002385 回文素数
	https://en.wikipedia.org/wiki/Palindromic_prime
https://oeis.org/A006567 反素数 emirp (primes whose reversal is a different prime)
	https://en.wikipedia.org/wiki/Emirp
https://oeis.org/A003459 绝对素数/可交换素数 Absolute primes (or permutable primes): every permutation of the digits is a prime
	https://en.wikipedia.org/wiki/Permutable_prime
https://oeis.org/A007500 Primes whose reversal in base 10 is also prime
https://oeis.org/A006995 二进制回文数
https://oeis.org/A007632 既是二进制回文数又是十进制回文数

https://oeis.org/A090994 Number of meaningful differential operations of the n-th order on the space R^9
a(k+5) = a(k+4) + 4*a(k+3) - 3*a(k+2) - 3*a(k+1) + a(k)
相关题目 LC1215 https://leetcode.cn/problems/stepping-numbers/

二进制字符串
https://oeis.org/A052944 a(n) = 2^n + n - 1  Shortest length of bit-string containing all bit-strings of given length n
https://math.stackexchange.com/questions/4509158/length-of-the-shortest-binary-string-that-contains-as-substrings-all-unique-n-le
https://en.wikipedia.org/wiki/De_Bruijn_sequence
应用 https://codeforces.com/problemset/problem/1469/E

套路题 https://codeforces.com/problemset/problem/1415/D
按位归纳 https://codeforces.com/problemset/problem/925/C

todo O(1) https://codeforces.com/contest/520/submission/205035892
*/

// Bitset
// 有时候也可以用 big.Int 代替
// 部分参考 C++ 的标准库源码 https://gcc.gnu.org/onlinedocs/libstdc++/libstdc++-html-USERS-3.4/bitset-source.html
// NOTE: 若要求方法内不修改 b 而是返回一个修改后的拷贝，可以在方法开头加上 b = append(Bitset(nil), b...) 并返回 b
// NOTE: 如果效率不够高，可以试试 0-1 线段树，见 segment_tree01.go
//
// https://codeforces.com/problemset/problem/33/D（也可以用 LCA）
// https://codeforces.com/contest/1826/problem/E
// https://atcoder.jp/contests/abc258/tasks/abc258_g
const _w = bits.UintSize

func NewBitset(n int) Bitset { return make(Bitset, n/_w+1) } // (n+_w-1)/_w

type Bitset []uint

func (b Bitset) Has(p int) bool { return b[p/_w]&(1<<(p%_w)) != 0 } // get
func (b Bitset) Flip(p int)     { b[p/_w] ^= 1 << (p % _w) }
func (b Bitset) Set(p int)      { b[p/_w] |= 1 << (p % _w) }  // 置 1
func (b Bitset) Reset(p int)    { b[p/_w] &^= 1 << (p % _w) } // 置 0

func (b Bitset) SetAll1() {
	for i := range b {
		b[i] = math.MaxUint
	}
}

// 遍历所有 1 的位置
// 如果对范围有要求，可在 f 中 return p < n
func (b Bitset) Foreach(f func(p int) (Break bool)) {
	for i, v := range b {
		for ; v > 0; v &= v - 1 {
			j := i*_w | bits.TrailingZeros(v)
			if f(j) {
				return
			}
		}
	}
}

// 返回第一个 0 的下标，若不存在则返回一个不小于 n 的位置
func (b Bitset) Index0() int {
	for i, v := range b {
		if ^v != 0 {
			return i*_w | bits.TrailingZeros(^v)
		}
	}
	return len(b) * _w
}

// 返回第一个 1 的下标，若不存在则返回一个不小于 n 的位置（同 C++ 中的 _Find_first）
func (b Bitset) Index1() int {
	for i, v := range b {
		if v != 0 {
			return i*_w | bits.TrailingZeros(v)
		}
	}
	return len(b) * _w
}

// 返回下标 >= p 的第一个 1 的下标，若不存在则返回一个不小于 n 的位置（类似 C++ 中的 _Find_next，这里是 >=）
func (b Bitset) Next1(p int) int {
	if i := p / _w; i < len(b) {
		v := b[i] & (^uint(0) << (p % _w)) // mask off bits below bound
		if v != 0 {
			return i*_w | bits.TrailingZeros(v)
		}
		for i++; i < len(b); i++ {
			if b[i] != 0 {
				return i*_w | bits.TrailingZeros(b[i])
			}
		}
	}
	return len(b) * _w
}

// 返回下标 >= p 的第一个 0 的下标，若不存在则返回一个不小于 n 的位置
func (b Bitset) Next0(p int) int {
	if i := p / _w; i < len(b) {
		v := b[i]
		if p%_w > 0 {
			v |= ^(^uint(0) << (p % _w))
		}
		if ^v != 0 {
			return i*_w | bits.TrailingZeros(^v)
		}
		for i++; i < len(b); i++ {
			if ^b[i] != 0 {
				return i*_w | bits.TrailingZeros(^b[i])
			}
		}
	}
	return len(b) * _w
}

// 返回最后第一个 1 的下标，若不存在则返回 -1
func (b Bitset) LastIndex1() int {
	for i := len(b) - 1; i >= 0; i-- {
		if b[i] != 0 {
			return i*_w | (bits.Len(b[i]) - 1) // 如果再 +1，需要改成 i*_w + bits.Len(b[i])
		}
	}
	return -1
}

// += 1 << i，模拟进位
func (b Bitset) Add(i int) { b.FlipRange(i, b.Next0(i)) }

// -= 1 << i，模拟借位
func (b Bitset) Sub(i int) { b.FlipRange(i, b.Next1(i)) }

// 判断 [l,r] 范围内的数是否全为 0
// https://codeforces.com/contest/1107/problem/D（标准做法是二维前缀和）
func (b Bitset) All0(l, r int) bool {
	i := l / _w
	if i == r/_w {
		mask := ^uint(0)<<(l%_w) ^ ^uint(0)<<(r%_w)
		return b[i]&mask == 0
	}
	if b[i]>>(l%_w) != 0 {
		return false
	}
	for i++; i < r/_w; i++ {
		if b[i] != 0 {
			return false
		}
	}
	mask := ^uint(0) << (r % _w)
	return b[r/_w]&^mask == 0
}

// 判断 [l,r] 范围内的数是否全为 1
func (b Bitset) All1(l, r int) bool {
	i := l / _w
	if i == r/_w {
		mask := ^uint(0)<<(l%_w) ^ ^uint(0)<<(r%_w)
		return b[i]&mask == mask
	}
	mask := ^uint(0) << (l % _w)
	if b[i]&mask != mask {
		return false
	}
	for i++; i < r/_w; i++ {
		if ^b[i] != 0 {
			return false
		}
	}
	mask = ^uint(0) << (r % _w)
	return ^(b[r/_w] | mask) == 0
}

// 反转 [l,r) 范围内的比特
// https://codeforces.com/contest/1705/problem/E
func (b Bitset) FlipRange(l, r int) {
	maskL, maskR := ^uint(0)<<(l%_w), ^uint(0)<<(r%_w)
	i := l / _w
	if i == r/_w {
		b[i] ^= maskL ^ maskR
		return
	}
	b[i] ^= maskL
	for i++; i < r/_w; i++ {
		b[i] = ^b[i]
	}
	b[i] ^= ^maskR
}

// 将 [l,r) 范围内的比特全部置 1
func (b Bitset) SetRange(l, r int) {
	maskL, maskR := ^uint(0)<<(l%_w), ^uint(0)<<(r%_w)
	i := l / _w
	if i == r/_w {
		b[i] |= maskL ^ maskR
		return
	}
	b[i] |= maskL
	for i++; i < r/_w; i++ {
		b[i] = ^uint(0)
	}
	b[i] |= ^maskR
}

// 将 [l,r) 范围内的比特全部置 0
func (b Bitset) ResetRange(l, r int) {
	maskL, maskR := ^uint(0)<<(l%_w), ^uint(0)<<(r%_w)
	i := l / _w
	if i == r/_w {
		b[i] &= ^maskL | maskR
		return
	}
	b[i] &= ^maskL
	for i++; i < r/_w; i++ {
		b[i] = 0
	}
	b[i] &= maskR
}

// 左移 k 位
// LC1981 https://leetcode.cn/problems/minimize-the-difference-between-target-and-chosen-elements/
func (b Bitset) Lsh(k int) {
	if k == 0 {
		return
	}
	shift, offset := k/_w, k%_w
	if shift >= len(b) {
		for i := range b {
			b[i] = 0
		}
		return
	}
	if offset == 0 {
		// Fast path
		copy(b[shift:], b)
	} else {
		for i := len(b) - 1; i > shift; i-- {
			b[i] = b[i-shift]<<offset | b[i-shift-1]>>(_w-offset)
		}
		b[shift] = b[0] << offset
	}
	for i := 0; i < shift; i++ {
		b[i] = 0
	}
}

// 右移 k 位
func (b Bitset) Rsh(k int) {
	if k == 0 {
		return
	}
	shift, offset := k/_w, k%_w
	if shift >= len(b) {
		for i := range b {
			b[i] = 0
		}
		return
	}
	lim := len(b) - 1 - shift
	if offset == 0 {
		// Fast path
		copy(b, b[shift:])
	} else {
		for i := 0; i < lim; i++ {
			b[i] = b[i+shift]>>offset | b[i+shift+1]<<(_w-offset)
		}
		// 注意：若前后调用 lsh 和 rsh，需要注意超出 n 的范围的 1 对结果的影响（如果需要，可以把范围开大点）
		b[lim] = b[len(b)-1] >> offset
	}
	for i := lim + 1; i < len(b); i++ {
		b[i] = 0
	}
}

// 借用 bits 库中的一些方法的名字
func (b Bitset) OnesCount() (c int) {
	for _, v := range b {
		c += bits.OnesCount(v)
	}
	return
}
func (b Bitset) TrailingZeros() int { return b.Index1() }
func (b Bitset) Len() int           { return b.LastIndex1() + 1 }

// 下面几个方法均需保证长度相同
func (b Bitset) Equals(c Bitset) bool {
	for i, v := range b {
		if v != c[i] {
			return false
		}
	}
	return true
}

func (b Bitset) HasSubset(c Bitset) bool {
	for i, v := range b {
		if v|c[i] != v {
			return false
		}
	}
	return true
}

// 将 c 的元素合并进 b
func (b Bitset) UnionFrom(c Bitset) {
	for i, v := range c {
		b[i] |= v
	}
}

func (b Bitset) IntersectionFrom(c Bitset) {
	for i, v := range c {
		b[i] &= v
	}
}

// 注：有关子集枚举的位运算技巧，见 search.go
func _(x int) {
	// 利用 -v = ^v+1
	lowbit := func(v int) int { return v & -v }

	// 最低位的 1 变 0
	x &= x - 1

	// 最低位的 0 变 1
	x |= x + 1

	// x 是 y 的子集
	isSubset := func(x, y int) bool { return x|y == y } // x 和 y 的并集是 y
	isSubset = func(x, y int) bool { return x&y == x }  // x 和 y 的交集是 x

	// 1,2,4,8,...
	isPow2 := func(v int) bool { return v > 0 && v&(v-1) == 0 }

	// 是否有两个相邻的 1    有 https://oeis.org/A004780 没有 https://oeis.org/A003714
	hasAdjacentOnes := func(v uint) bool { return v>>1&v > 0 }

	// 是否有两个相邻的 0（不考虑前导零）    有 https://oeis.org/A004753 没有 https://oeis.org/A003754
	hasAdjacentZeros := func(v uint) bool {
		v |= v >> 1 // 若没有相邻的 0，则 v 会变成全 1 的数
		return v&(v+1) > 0
	}

	bits31 := func(v int) []byte {
		bits := make([]byte, 31)
		for i := range bits {
			bits[i] = byte(v >> (30 - i) & 1)
		}
		return bits
	}
	_bits31 := func(v int) string { return Sprintf("%031b", v) }
	_bits32 := func(v uint) string { return Sprintf("%032b", v) }

	// https://www.acwing.com/problem/content/293/
	initEvenZeros := func(n int) {
		mask := 1 << n
		// 在 i 的长为 n 二进制表示中，如果所有连续 0 的个数都是偶数（包括前导零），则 evenZeros[i] 为 true，否则为 false
		evenZeros := make([]bool, mask)
	next:
		for i := range evenZeros {
			for s, pre := uint(mask|i), -1; s > 0; s &= s - 1 {
				p := bits.TrailingZeros(s)
				if (p-pre)%2 == 0 { // 开区间 (pre,p) 中有奇数个连续 0
					continue next
				}
				pre = p
			}
			evenZeros[i] = true
		}
	}

	// 返回最小的非负 x，其满足 n^x >= m
	// https://codeforces.com/problemset/problem/1554/C
	leastXor := func(n, m int) (res int) {
		for i := 29; i >= 0; i-- { // 29 for 1e9
			bn, bm := n>>i&1, m>>i&1
			if bn == 1 && bm == 0 { // 后面都填 0
				break
			}
			if bn == 0 && bm == 1 { // 必须填 1
				res |= 1 << i
			}
			// bn = bm 的情况填 0
		}
		return
	}

	// 对于数组 a 的所有区间，返回 op(区间元素) 的全部运算结果    logTrick
	// 利用操作的单调性求解
	// 时间复杂度：O(fnlogU)，其中 f 为 op(x,y) 的时间复杂度，一般是 O(1)，n=len(a)，U=max(a)
	// 空间复杂度：O(logU)，返回值不计入
	// |: LC898 https://leetcode.cn/problems/bitwise-ors-of-subarrays/ 2133
	//    - 原题 https://codeforces.com/problemset/problem/243/A 1600
	//    LC2411 https://leetcode.cn/problems/smallest-subarrays-with-maximum-bitwise-or/ 1938
	//    https://www.luogu.com.cn/problem/P8569
	//    - 做法见下面的 bitOpTrickCnt      
	//    - 题目源于这场比赛 https://www.luogu.com.cn/contest/65460#problems
	//    - 其它做法 https://www.luogu.com.cn/blog/203623/sol-The-seventh-district
	// &: LC1521 https://leetcode.cn/problems/find-a-value-of-a-mysterious-function-closest-to-target/
	// GCD: 原理：固定右端点时，向左扩展，GCD 要么不变，要么至少减半，所以固定右端点时，只有 O(log U) 个 GCD
	//      LC2447 https://leetcode.cn/problems/number-of-subarrays-with-gcd-equal-to-k/
	//      LC2654 https://leetcode.cn/problems/minimum-number-of-operations-to-make-all-array-elements-equal-to-1/ 
	//             https://www.dotcpp.com/oj/problem2709.html
	//      LC2941 https://leetcode.cn/problems/maximum-gcd-sum-of-a-subarray/
	//      https://codeforces.com/edu/course/2/lesson/9/2/practice/contest/307093/problem/G
	//      https://codeforces.com/problemset/problem/891/A
	//      https://codeforces.com/problemset/problem/475/D (见下面的 bitOpTrickCnt)
	//      https://codeforces.com/problemset/problem/1632/D (见下面的 bitOpTrickCnt)
	//      已知所有 GCD 还原数组 a https://codeforces.com/problemset/problem/894/C
	//      (子数组长度 * 子数组 GCD) 的最大值 https://www.luogu.com.cn/problem/P5502
	// LCM: LC2470 https://leetcode.cn/problems/number-of-subarrays-with-lcm-equal-to-k/
	//      https://codeforces.com/contest/1834/problem/E
	bitOpTrick := func(a []int, op func(x, y int) int) map[int]struct{} {
		ans := map[int]struct{}{}
		opRes := []int{} // 以 a[i] 结尾的所有子数组，在 op 操作下的所有结果
		for _, v := range a {
			for j, x := range opRes {
				opRes[j] = op(x, v) // 每个都和新遍历到的 a[i] 算一下
			}
			opRes = append(opRes, v)      // a[i] 单独组成一个子数组
			opRes = slices.Compact(opRes) // 去重
			for _, w := range opRes {
				// 统计 w... 根据题目改动
				ans[w] = struct{}{}
			}
		}
		return ans
	}

	// 进阶：对于数组 a 的所有区间，返回 op(区间元素) 的全部运算结果及其出现次数
	// 甚至还可以做到把每个运算结果对应的每个区间长度的出现次数求出来（需要差分）
	// https://codeforces.com/problemset/problem/475/D
	// https://codeforces.com/problemset/problem/1632/D
	// 与单调栈结合 https://codeforces.com/problemset/problem/875/D
	// CERC13，紫书例题 10-29，UVa 1642 https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=825&page=show_problem&problem=4517
	// https://www.luogu.com.cn/problem/P8569
	bitOpTrickCnt := func(a []int, op func(x, y int) int) map[int]int {
		cnt := map[int]int{}
		// 视情况，r 可以省略
		type result struct{ v, l, r int } // [l,r)
		opRes := []result{}
		for i, v := range a {
			for j, p := range opRes {
				opRes[j].v = op(p.v, v)
			}
			opRes = append(opRes, result{v, i, i + 1})
			// 去重（合并 v 相同的 result）
			j := 0
			for _, q := range opRes[1:] {
				if opRes[j].v != q.v {
					j++
					opRes[j] = q
				} else {
					opRes[j].r = q.r // 如果省略 r 的话，这行可以去掉
				}
			}
			opRes = opRes[:j+1]
			// 此时我们将区间 [0,i] 划分成了 len(set) 个左闭右开区间
			// 对 ∀p∈set，∀j∈[p.l,p.r)，op(区间[j,i]) 的计算结果均为 p.v
			for _, p := range opRes {
				// do p...     [l,r)
				cnt[p.v] += p.r - p.l
			}
		}
		return cnt
	}

	//（接上）考虑乘法
	// 输入：数组 a，元素均为正整数
	// 输出：满足【元素和等于元素乘积】的非空连续子数组的个数
	// 我们来考虑对每个区间右端点，有多少个合法的区间左端点
	// 核心思路是，对于每个满足题目要求的区间，其区间积不会超过 sum(a)
	// 由于乘积至少要乘 2 才会变化，所以对于一个固定的区间右端点，不同的区间积至多有 O(log(sum(a))) 个
	// 同时由于元素均为正数，所以对一个固定的区间右端点，区间左端点也至多有 O(log(sum(a))) 个
	// 据此我们只需要在加入一个新的数后，去重并去掉区间积超过 sum(a) 的区间，就可以暴力做出此题
	// 注：根据以上推导过程，我们还可以得出总的答案个数至多为 O(nlog(sum(a)))
	// https://www.dotcpp.com/oj/problem2622.html
	// https://codeforces.com/problemset/problem/1872/G 2000
	// 变形·面试题：把「区间和」改成「区间异或和」
	countSumEqMul := func(a []int) (ans int) {
		tot := 0
		for _, v := range a {
			tot += v
		}
		// 每个前缀和互不相同
		posS := map[int]int{0: 0}
		sum := 0
		type result struct{ v, l, r int }
		muls := []result{}
		for i, v := range a {
			sum += v
			for j := range muls {
				muls[j].v *= v
			}
			muls = append(muls, result{v, i, i + 1})
			// 去重
			j := 0
			for _, q := range muls[1:] {
				if muls[j].v != q.v {
					j++
					muls[j] = q
				} else {
					muls[j].r = q.r
				}
			}
			muls = muls[:j+1]
			// 去掉超过 tot 的，从而保证 muls 中至多有 O(log(tot)) 个元素
			for muls[0].v > tot {
				muls = muls[1:]
			}
			// 此时我们将区间 [muls[0].l,i] 划分成了 len(muls) 个（左闭右开）区间，对 ∀j∈[muls[k].l,muls[k].r)，[j,i] 的区间积均为 muls[k].v
			for _, p := range muls {
				// 判断左端点前缀和对应下标是否在范围内
				if pos, has := posS[sum-p.v]; has && p.l <= pos && pos < p.r {
					ans++
				}
			}
			posS[sum] = i + 1
		}
		return
	}

	// 找三个不同的在 [l,r] 范围内的数，其异或和为 0
	// 考虑尽可能地小化最大减最小的值，构造 (x, y, z) = (b*2-1, b*3-1, b*3), b=2^k
	// 相关题目 https://codeforces.com/problemset/problem/460/D
	zeroXorSum3 := func(l, r int) []int {
		for b := 1; b*3 <= r; b <<= 1 {
			if x, y, z := b*2-1, b*3-1, b*3; l <= x && z <= r {
				return []int{x, y, z}
			}
		}
		return nil
	}

	// 在 [low,high] 区间内找两个数字 A B，使其异或值最大且不超过 limit
	// 返回值保证 A <= B
	// 复杂度 O(log(high))
	maxXorWithLimit := func(low, high, limit int) (int, int) {
		n := bits.Len(uint(high ^ low))
		maxXor := 1<<n - 1
		mid := high&^maxXor | 1<<(n-1)
		if limit >= maxXor { // 无约束，相关题目 https://codeforces.com/problemset/problem/276/D
			return mid - 1, mid
		}
		if limit >= 1<<(n-1) { // A 和 B 能否在第 n-1 位不同的情况下，构造出一个满足要求的解？
			a, b := mid&(mid-1), mid
			for i := n - 2; i >= 0; i-- {
				bt := 1 << i
				if limit&bt > 0 { // a 取 1，b 取 0 总是优于 a 取 0，b 取 1
					a |= bt
				} else if high&(bt<<1-1) > ^low&(bt<<1-1) { // high 侧大，都取 1
					if high&bt == 0 { // b 没法取 1
						goto next
					}
					a |= bt
					b |= bt
				} else {            // low 侧大，都取 0
					if low&bt > 0 { // a 没法取 0
						goto next
					}
				}
				if (a^low)&bt > 0 { // a 不受 low 的约束
					a |= limit & (bt - 1)
					break
				}
				if (b^high)&bt > 0 { // b 不受 high 的约束
					a |= bt - 1
					b |= ^limit & (bt - 1)
					break
				}
			}
			return a, b
		}
		// A 和 B 在第 n-1 位上必须相同
	next:
		f := func(high int) (int, int) {
			n := bits.Len(uint(high ^ mid))
			maxXor := min(1<<n-1, limit)
			// 只有当 maxXor 为 0 时，返回值才必须相等
			if maxXor == 0 {
				return mid, mid
			}
			// maxXor 的最高位置于 B，其余置于 A
			mb := 1 << (bits.Len(uint(maxXor)) - 1)
			return mid | maxXor&^mb, mid | mb
		}
		if high-mid > mid-1-low { // 选区间长的一侧
			return f(high)
		}
		a, b := f(2*mid - 1 - low) // 对称到 high
		return 2*mid - 1 - b, 2*mid - 1 - a
	}

	_ = []interface{}{
		lowbit, isSubset, isPow2, hasAdjacentOnes, hasAdjacentZeros, bits31, _bits31, _bits32, initEvenZeros,
		leastXor,
		bitOpTrick, bitOpTrickCnt, countSumEqMul,
		zeroXorSum3,
		maxXorWithLimit,
	}
}

// https://halfrost.com/go_s2_de_bruijn/

// LC137 https://leetcode.cn/problems/single-number-ii/
// 除了某个元素只出现一次以外，其余每个元素均出现了三次。返回只出现了一次的元素
// 		定义两个集合 ones 和 twos，初始为空
// 		第一次出现就放在 ones 中
//		第二次出现就在 ones 中删除并放在 twos
//		第三次出现就从 twos 中删除
//		这样最终 ones 中就留下了最后的结果
func singleNumber(a []int) int {
	ones, twos := 0, 0
	for _, v := range a {
		ones = (ones ^ v) &^ twos
		twos = (twos ^ v) &^ ones
	}
	return ones
}