日期:2014-05-20  浏览次数:20866 次

在Java区看到一个比较有意思的题目,转过来大家讨论一下
Consider all integer combinations of a^b for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:

  2^2=4, 2^3=8, 2^4=16, 2^5=32
  3^2=9, 3^3=27, 3^4=81, 3^5=243
  4^2=16, 4^3=64, 4^4=256, 4^5=1024
  5^2=25, 5^3=125, 5^4=625, 5^5=3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

翻译一下

就是有一个m*n的矩阵,

该矩阵的第一列是a^b,(a+1)^b,.....(a + n - 1)^b
第二列是a^(b+1),(a+1)^(b+1),.....(a + n - 1)^(b+1)
.......
第m列是(b + m - 1),(a+1)^(b + m - 1),.....(a + n - 1)^(b + m - 1)

问这个矩阵里有多少不重复的数
(比如4^3 = 8^2,这样的话就有重复了)

原题里面m和n的范围为100,这样的话就不用优化了,为了把问题搞难一些,让m和n都为10000吧。

------解决方案--------------------
让我想想。。。

因数分解?


不好意思。。。我的程序非常烂的。

C# code
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        class Node
        {
            public int Num { get; set; }
            public Dictionary<int, int> Count { get; set; }
            public Node() { Count = new Dictionary<int, int>(); }
            public override string ToString()
            {
                string s = "";
                foreach (var i in Count)
                {
                    s += "|" + Math.Pow((double)Num, (double)i.Key).ToString() + " " + i.Value.ToString();
                }
                return s;
            }
        }

        static Dictionary<int, Node> Nodes = new Dictionary<int, Node>();

        static List<int> Primers = new List<int>() { };

        static void ProcPrimes(int num)
        {
            bool noprime = false;
            foreach (var i in Primers)
            {
                if (num % i == 0)
                {
                    noprime = true;
                    break;
                }
                else
                {
                    if (i > num) break;
                }
            }
            if (!noprime) Primers.Add(num);
        }

        static void ToNodes(int num, int power)
        {
            List<int> list = new List<int>();
            foreach (var i in Primers)
            {
                if (num % i == 0)
                {
                    list.Add(i);
                    break;
                }
                else
                {
                    if (i > num) break;
                }
            }
            int x = 1;
            list.ForEach(x1 => x *= x1);
            if (!Nodes.ContainsKey(x)) Nodes.Add(x, new Node(){ Num = x });
            int pwr = num / x * power;
            if (!Nodes[x].Count.ContainsKey(pwr)) 
                Nodes[x].Count.Add(pwr, 1);
            else
                Nodes[x].Count[pwr]++;
        }

        static int Solve(int startA, int endA, int startB, int endB)
        {
            Debug.Assert(startA > 1 && startB > 1);
            Debug.Assert(endA > startA && endB > startB);

            for (int i = 2; i <= endA; i++)
                ProcPrimes(i);
            for (int i = startA; i <= endA; i++)
                for (int j = startB; j <= endB; j++)
                    ToNodes(i, j);

            Nodes.Values.ToList()
                .ForEach(x => (from y
                               in x.Count
                               where y.Value > 0
                               select Math.Pow((double)x.Num, (double)y.Key))
                                .ToList().ForEach(z => Debug.WriteLine(z)));

            int n = 0;
            Nodes.ToList().ForEach(x => n += x.Value.Count.Count);
            return n;            
        }

        static void Main(string[] args)
        {
            Console.WriteLine(Solve(2, 1000, 2, 1000));
        }
        
    }
}