从给定范围中删除给定数字集的所有倍数

Posted 2021-04-15

我坚持一个问题，它说，给出一个数字N和一组数字，S = {s1,s2,.....sn}其中s1 < s2 < sn < N，从{s1, s2,....sn}范围中删除1..N的所有倍数

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 

我想有一个算法方法，伪代码而不是完整的解决方案。

我尝试过的：

(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！

答案

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 / s1次，然后对于第二个元素执行N / s2，依此类推。

我们需要估计N / s1 + N / s2 + ... + N / sn的大小。

N / s1 + N / s2 + ... + N / sn = N *（1 / s1 + 1 / s2 + ... + 1 / sn）<= N *（1/1 + 1/2 + .. 。+ 1 / n）。

最后的不等式是由于s1 <s2 <... <sn的事实，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，（例如参见this），因此上述算法的时间复杂度为O（N log（ N））。

另一答案

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

另一答案

由于S被排序，我们可以通过跳过已标记的S（O(N)）中的元素来保证http://codepad.org/Joflhb7x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)