来一个编程题吧

[TheMatrix]

这个问题不要求效率的话，在leetcode里也是hard的。

问题的大意是：以质因数分解的形式遍历正整数。

要求这样遍历：以总指数和参与的质数的个数同步增长的方式遍历。

比如考虑20以内的质数：[2, 3, 5, 7, 11, 13, 17, 19]。一共8个。这些是参与的质数的个数。那么总指数小于等于8，可以叫level。

比如：

level=1，有[2⁰,2¹]，两个。

level=2，有[2⁰3⁰,2¹3⁰,2⁰3¹,2¹3¹,2²3⁰,2⁰3²]，共6个。

level=3，有[2⁰3⁰5⁰,2¹3⁰5⁰,...]，共...我也不知道多少个。

结果可以列成序列的形式：
(2,0,0,3) - 这代表2²7³。

能去掉duplicate最好。

[YWY]

level 1: 2⁰, 2¹; total number is 2 = 2¹.
level 2: 2ⁱ3^j with i, j in {0, 1, 2}; total number is 9 = 3² with duplicates from level 1; removing level 1, there are 3² - 2¹ numbers.
...
level n: there are (n+1)ⁿ numbers, including numbers from lower levels; if we removed the numbers from lower levels, there are (n+1)ⁿ - n^n-1 numbers.

[TheMatrix]

level 1: 2⁰, 2¹; total number is 2 = 2¹.
level 2: 2ⁱ3^j with i, j in {0, 1, 2}; total number is 9 = 3² with duplicates from level 1; removing level 1, there are 3² - 2¹ numbers.
...
level n: there are (n+1)ⁿ numbers, including numbers from lower levels; if we removed the numbers from lower levels, there are (n+1)ⁿ - n^n-1 numbers.

level 2应该是6个。

[YWY]

level 2应该是6个。

Level 2: 2⁰3⁰, 2¹3⁰, 2²3⁰, 2⁰3¹, 2¹3¹, 2²3¹, 2⁰3², 2¹3², 2²3². The first 2 numbers, 2⁰3⁰ and 2¹3⁰, are from level 1.

[TheMatrix]

Level 2: 2⁰3⁰, 2¹3⁰, 2²3⁰, 2⁰3¹, 2¹3¹, 2²3¹, 2⁰3², 2¹3², 2²3². The first 2 numbers, 2⁰3⁰ and 2¹3⁰, are from level 1.

哦。我说的是总指数。指数相加。

[YWY]

哦。我说的是总指数。指数相加。

At level n, we need to find the total number of monomials in n variables with total degree up to n. This number is known to be "2n-choose-n". This allows duplicates from lower levels.

Note that each level is contained in the next level (i.e., level i is contained in level i+1). If we remove lower levels from level n, the answer is 2n-choose-n minus (2n-2)-choose-(n-1).

[TheMatrix]

At level n, we need to find the total number of monomials in n variables with total degree up to n. This number is known to be "2n-choose-n". This allows duplicates from lower levels.

Note that each level is contained in the next level (i.e., level i is contained in level i+1). If we remove lower levels from level n, the answer is 2n-choose-n minus (2n-2)-choose-(n-1).

这是对的。

怎么证明呢？

[YWY]

这是对的。

怎么证明呢？

https://en.wikipedia.org/wiki/Monomial

https://en.wikipedia.org/wiki/Stars_and ... atorics%29

[TheMatrix]

https://en.wikipedia.org/wiki/Monomial

https://en.wikipedia.org/wiki/Stars_and ... atorics%29

嗯。这个问题我以前也问过。有人给过方法。nonnegative的更绕一些。

[TheMatrix]

这是个编程题。我们还是上个程序吧。

这个问题抽象出来之后，没有我想象的难。和素数没有什么关系。主要是个排列组合的问题。

不过用到递归的话，在leetcode里应该都算hard的。

代码： 全选

def level_n(N):
    result = [[]]
    for nstep in range(N):
        nresult = []
        for path in result:
            s = sum(path)
            nresult += [path+[i] for i in range(N-s+1)]
        result = nresult
    return result

[Dahuaidanyimei]

抱歉我懒，直接copy 进了 copilot，结果我也没验证：



def prime_factors(n):
factors = []
for i in range(2, int(n**0.5) + 1):
while n % i == 0:
factors.append(i)
n //= i
if n > 1:
factors.append(n)
return factors

def count_prime_factors(n):
factors = prime_factors(n)
unique_factors = set(factors)
return len(unique_factors)

def generate_sequence(limit):
sequence = []
for i in range(1, limit + 1):
total_exponent = count_prime_factors(i)
num_unique_factors = count_prime_factors(total_exponent)
sequence.append((total_exponent, num_unique_factors))
return sequence

# Example usage for 20:
result = generate_sequence(20)
print(result)

[TheMatrix]

抱歉我懒，直接copy 进了 copilot，结果我也没验证：



def prime_factors(n):
factors = []
for i in range(2, int(n**0.5) + 1):
while n % i == 0:
factors.append(i)
n //= i
if n > 1:
factors.append(n)
return factors

def count_prime_factors(n):
factors = prime_factors(n)
unique_factors = set(factors)
return len(unique_factors)

def generate_sequence(limit):
sequence = []
for i in range(1, limit + 1):
total_exponent = count_prime_factors(i)
num_unique_factors = count_prime_factors(total_exponent)
sequence.append((total_exponent, num_unique_factors))
return sequence

# Example usage for 20:
result = generate_sequence(20)
print(result)

能运行 - 调整一下缩进 - 这就不错了。