Question

我遇到了一个问题，它给出了一个N数字和一组数字，S = {s1,s2,.....sn} s1 < s2 < sn < N，删除{s1, s2,....sn}的所有倍数来自范围1..N

示例：

Let N = 10
S = {2,4,5}
Output: {1, 7, 9}
Explanation: multiples of 2 within range: 2, 4, 6, 8
             multiples of 4 within range: 4, 8
             multiples of 5 within range: 5, 10

我想有一种算法方法，即伪代码而不是完整的解决方案。

我尝试了什么：

(Considering the same example as above) 

 1. For the given N, find all the prime factors of that number.
    Therefore, for 10, prime-factors are: 2,3,5,7
    In the given set, S = {2,4,5}, the prime-factors missing from 
    {2,3,5,7} are {3,7}.  
 2. First, check prime-factors that are present: {2,5}
    Hence, all the multiples of them will be removed 
    {2,4,5,6,8,10}
 3. Check for non-prime numbers in S = {4}
 4. Check, if any divisor of these numbers has already been 
    previously processed.
         > In this case, 2 is already processed.
         > Hence no need to process 4, as all the multiples of 4 
           would have been implicitly checked by the previous 
           divisor.
    If not,
         > Remove all the multiples from this range.
 5. Repeat for all the remaining non primes in the set.

请提出您的想法！

Answer 1

可以使用类似于Eratosthenes筛子的东西在O（N log（n））时间和O（N）额外内存中解决它。

isMultiple[1..N] = false

for each s in S:
    t = s
    while t <= N:
        isMultiple[t] = true
        t += s

for i in 1..N:
    if not isMultiple[i]:
        print i

这使用O（N）内存来存储isMultiple数组。

时间复杂度为O（N log（n））。实际上，对于S中的第一个元素，内部while循环将执行N / s ₁次，然后对于第二个元素执行N / s ₂，依此类推。

我们需要估计N / s ₁ + N / s ₂ + ... + N / s _n的幅度。

N / s ₁ + N / s ₂ + ... + N / s _n = N *（1 / s ₁ + 1 / s ₂ + ... + 1 / s _n）＆lt; = N *（ 1/1 + 1/2 + ... + 1 / n）。

最后的不等式是由于s ₁＆lt; s ₂＆lt; ... ＆LT; s _n，因此最坏的情况是它们取值{1,2，... n}。

然而，谐波系列1/1 + 1/2 + ... + 1 / n在O（log（n））中，（例如见this），因此，上述算法的时间复杂度为O（N log（n））。

Answer 2

基本解决方案：

let set X be our output set.

for each number, n, between 1 and N:
    for each number, s, in set S:
        if s divides n:
            stop searching S, and move onto the next number,n.
        else if s is the last element in S:
            add n to the set X.

你可以在运行这个算法之前显然删除S中的倍数，但我不认为素数是可行的方法

Answer 3

由于S已排序，我们可以通过跳过已标记为（http://codepad.org/Joflhb7x）的S中的元素来保证O(N)的复杂性：

N = 10
S = [2,4,5]
marked = set()
i = 0
curr = 1

while curr <= N:
  while curr < S[i]:
    print curr
    curr = curr + 1

  if not S[i] in marked:
    mult = S[i]

    while mult <= N:
      marked.add(mult)
      mult = mult + S[i]

  i = i + 1
  curr = curr + 1

  if i == len(S):
    while curr <= N:
      if curr not in marked:
        print curr
      curr = curr + 1

print list(marked)

从给定范围中删除给定数字集的所有倍数

3 个答案: