假設你不想使用LINQ，下面是一個實現方法。這表示它可怕地未被優化，並且我建議使用更短的LINQ實現。 （見：Marc Gravell's post。）
using System;
using System.Collections.Generic;
namespace MathIterator
{
class Program
{
static readonly int[] _inputs = new int[] { 5, 5, 5, 5 };
static readonly char[] _operations = new char[] { '+', '-', '*', '/' };
static Dictionary<int, List<string>> _calculations = new Dictionary<int, List<string>>();
static void Main(string[] args)
{
StartPermutation();
PrintResults();
}
static void StartPermutation()
{
if (_inputs.Length > 0)
Permute(1 /*index*/, _inputs[0], _inputs[0].ToString());
}
static void Permute(int index, int result, string computation)
{
if (index == _inputs.Length)
{
if (!_calculations.ContainsKey(result))
{
_calculations[result] = new List<string>();
}
_calculations[result].Add(computation);
}
else
{
foreach (char operation in _operations)
{
string nextComputation = String.Format("({0} {1} {2})",computation, operation, _inputs[index]);
int nextResult = result;
switch (operation)
{
case '+':
nextResult += _inputs[index];
break;
case '-':
nextResult -= _inputs[index];
break;
case '*':
nextResult *= _inputs[index];
break;
case '/':
nextResult /= _inputs[index];
break;
}
Permute(
index + 1,
nextResult,
nextComputation);
}
}
}
static void PrintResults()
{
for (int i = 1; i <= 10; ++i)
{
if (_calculations.ContainsKey(i))
{
Console.WriteLine("Found {0} entries for key {1}", _calculations[i].Count, i);
foreach (string calculation in _calculations[i])
{
Console.WriteLine(i + " = " + calculation);
}
}
else
{
Console.WriteLine("No entry for key: " + i);
}
}
}
}
}
這是另一種實現。這一個按照優先順序排列。再次，我不建議像這樣解決它。更重要的是，現在給予廣泛的短跑黑客（以區別於減號），但它似乎工作。
using System;
using System.Collections.Generic;
using System.Text.RegularExpressions;
namespace MathIterator
{
class Program
{
static readonly int[] _inputs = new[] { 5, 5, 5, 5 };
//HUGE hack, the '–' is a wide dash NOT a hyphen.
static readonly char[][] _operationLevels = new[] { new[] { '*', '/' }, new[] { '+', '–' } };
static List<string> _calculations = new List<string>();
static Dictionary<int, List<string>> _results = new Dictionary<int, List<string>>();
static void Main(string[] args)
{
StartPermutation();
StartEvaluateCalculations();
PrintResults();
}
static void StartPermutation()
{
if (_inputs.Length > 0)
Permute(1 /*index*/, _inputs[0].ToString());
}
static void Permute(int index, string computation)
{
if (index == _inputs.Length)
{
_calculations.Add(computation);
}
else
{
foreach (char[] operationLevel in _operationLevels)
{
foreach (char operation in operationLevel)
{
string nextComputation = String.Format("{0} {1} {2}", computation, operation, _inputs[index]);
Permute(
index + 1,
nextComputation);
}
}
}
}
static void StartEvaluateCalculations()
{
foreach (string calculation in _calculations)
{
int? result = EvaluateCalculation(calculation);
if (result != null)
{
int intResult = (int) result;
if (!_results.ContainsKey(intResult))
{
_results[intResult] = new List<string>();
}
_results[intResult].Add(calculation);
}
}
}
static int? EvaluateCalculation(string calculation)
{
foreach (char[] operationLevel in _operationLevels)
{
string[] results = calculation.Split(operationLevel, 2);
if (results.Length == 2)
{
int hitIndex = results[0].Length;
Regex firstDigit = new Regex(@"^ -?\d+");
Regex lastDigit = new Regex(@"-?\d+ $");
string firstMatch = lastDigit.Match(results[0]).Value;
int arg1 = int.Parse(firstMatch);
string lastMatch = firstDigit.Match(results[1]).Value;
int arg2 = int.Parse(lastMatch);
int result = 0;
switch (calculation[hitIndex])
{
case '+':
result = arg1 + arg2;
break;
//HUGE hack, the '–' is a wide dash NOT a hyphen.
case '–':
result = arg1 - arg2;
break;
case '*':
result = arg1 * arg2;
break;
case '/':
if ((arg2 != 0) && ((arg1 % arg2) == 0))
{
result = arg1/arg2;
break;
}
else
{
return null;
}
}
string prePiece = calculation.Remove(hitIndex - 1 - arg1.ToString().Length);
string postPiece = calculation.Substring(hitIndex + 1 + lastMatch.ToLower().Length);
string nextCalculation = prePiece + result + postPiece;
return EvaluateCalculation(nextCalculation);
}
}
return int.Parse(calculation);
}
static void PrintResults()
{
for (int i = 1; i <= 10; ++i)
{
if (_results.ContainsKey(i))
{
Console.WriteLine("Found {0} entries for key {1}", _results[i].Count, i);
foreach (string calculation in _results[i])
{
Console.WriteLine(i + " = " + calculation);
}
}
else
{
Console.WriteLine("No entry for key: " + i);
}
}
}
}
}
因此，基本上，你是通過所有可能的排列迭代和挑出那些給出正確答案的那些代碼並構建代表算術的字符串？小心分享您如何使用LINQ做到這一點？ – 2009-10-18 09:38:10
只是出於好奇：有沒有人設法生成3..8？我使用PowerShell進行的小測試只產生了1,2,9和10的結果。 – Joey 2009-10-18 09:50:45
是否允許使用括號？ （5 * 5 + 5）/ 5 == 6，（5 + 5 + 5）/ 5 == 3 – andyp 2009-10-18 10:08:41