如何加快我的Amicable数字算法？

Question

完成100,000的限制（n）需要很长时间。

我怀疑问题在于 CalculateAmicable（），其中数字越大，计算时间就越长。我可以改变什么来使它加速比这更快？

public static void Main (string[] args)
    {
        CheckAmicable (1, 100000);

    }

    public static int CheckAmicable(int start, int end)
    {
        for (int i = start; i < end; i++) {

            int main = CalculateAmicable (i); //220
            int temp = CalculateAmicable (main); //284
            int compare = CalculateAmicable (temp); //220
            if (compare == main) {
                if (main != temp && temp == i) {
                    Console.WriteLine (main + " = " + temp + " = " + compare + " i: " + i);
                    i = compare + 1;
                }
            }
        }
        return 0;
    }

    public static int CalculateAmicable(int number)
    {
        int total = 0;
        for (int i = 1; i < number; i++) {
            if (number%i == 0){
                total += i;
            }
        }
        return total;
    }

注意：英语不是我的第一语言！

Answer 1

是的，CalculateAmicable 效率低下 - O(N)复杂性 - 所以你有N次测试1e5 * 1e5 == 1e10 - 十亿/ em>操作很慢。 将复杂性降低到O(log(N))的方案是

  int n = 100000;

  //TODO: implement IList<int> GetPrimesUpTo(int) yourself
  var primes = GetPrimesUpTo((int)(Math.Sqrt(n + 1) + 1));

  // Key - number itself, Value - divisors' sum 
  var direct = Enumerable
    .Range(1, n)
    .AsParallel()
    .ToDictionary(index => index, 
                  index => GetDivisorsSum(index, primes) - index);

  var result = Enumerable
    .Range(1, n)
    .Where(x => x < direct[x] && 
                direct.ContainsKey((int) direct[x]) &&
                direct[(int) direct[x]] == x)
    .Select(x => $"{x,5}, {direct[x],5}");

  Console.Write(string.Join(Environment.NewLine, result));

结果在一秒之内（Core i7 3.2Ghz .Net 4.6 IA-64）：

  220,   284
 1184,  1210
 2620,  2924
 5020,  5564
 6232,  6368
10744, 10856
12285, 14595
17296, 18416
63020, 76084
66928, 66992
67095, 71145
69615, 87633
79750, 88730

详情GetDivisorsSum：

private static long GetDivisorsSum(long value, IList<int> primes) {
  HashSet<long> hs = new HashSet<long>();

  IList<long> divisors = GetPrimeDivisors(value, primes);

  ulong n = (ulong) 1;
  n = n << divisors.Count;

  long result = 1;

  for (ulong i = 1; i < n; ++i) {
    ulong v = i;
    long p = 1;

    for (int j = 0; j < divisors.Count; ++j) {
      if ((v % 2) != 0)
        p *= divisors[j];

      v = v / 2;
    }

    if (hs.Contains(p))
      continue;

    result += p;

    hs.Add(p);
  }

  return result;
}

GetPrimeDivisors：

private static IList<long> GetPrimeDivisors(long value, IList<int> primes) {
  List<long> results = new List<long>();

  int v = 0;
  long threshould = (long) (Math.Sqrt(value) + 1);

  for (int i = 0; i < primes.Count; ++i) {
    v = primes[i];

    if (v > threshould)
      break;

    if ((value % v) != 0)
      continue;

    while ((value % v) == 0) {
      value = value / v;

      results.Add(v);
    }

    threshould = (long) (Math.Sqrt(value) + 1);
  }

  if (value > 1)
    results.Add(value);

  return results;
}