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using System; | ||
using System.Collections.Generic; | ||
using System.Linq; | ||
using System.Text; | ||
using Ckknight.ProjectEuler.Collections; | ||
|
||
namespace Ckknight.ProjectEuler.Problems | ||
{ | ||
[Problem(55, | ||
@"If we take 47, reverse and add, 47 + 74 = 121, which is palindromic. | ||
Not all numbers produce palindromes so quickly. For example, | ||
349 + 943 = 1292, | ||
1292 + 2921 = 4213 | ||
4213 + 3124 = 7337 | ||
That is, 349 took three iterations to arrive at a palindrome. | ||
Although no one has proved it yet, it is thought that some numbers, | ||
like 196, never produce a palindrome. A number that never forms a | ||
palindrome through the reverse and add process is called a Lychrel | ||
number. Due to the theoretical nature of these numbers, and for the | ||
purpose of this problem, we shall assume that a number is Lychrel until | ||
proven otherwise. In addition you are given that for every number below | ||
ten-thousand, it will either (i) become a palindrome in less than fifty | ||
iterations, or, (ii) no one, with all the computing power that exists, | ||
has managed so far to map it to a palindrome. In fact, 10677 is the | ||
first number to be shown to require over fifty iterations before | ||
producing a palindrome: 4668731596684224866951378664 (53 iterations, | ||
28-digits). | ||
Surprisingly, there are palindromic numbers that are themselves Lychrel | ||
numbers; the first example is 4994. | ||
How many Lychrel numbers are there below ten-thousand? | ||
NOTE: Wording was modified slightly on 24 April 2007 to emphasise the | ||
theoretical nature of Lychrel numbers.")] | ||
public class Problem055 : BaseProblem | ||
{ | ||
public override object CalculateResult() | ||
{ | ||
return new Range(1, 10000) | ||
.Count(i => !new Range(50) | ||
.SelectWithAggregate(MathUtilities.ToDigits(i), (x, n) => | ||
{ | ||
var digits = x.Zip(x.Reverse(), (a, b) => a + b).ToList(); | ||
int carry = 0; | ||
for (int j = 0; j < digits.Count; j++) | ||
{ | ||
int current = digits[j] + carry; | ||
digits[j] = current % 10; | ||
carry = current / 10; | ||
} | ||
while (carry > 0) | ||
{ | ||
digits.Add(carry % 10); | ||
carry /= 10; | ||
} | ||
return digits.ToArray(); | ||
}) | ||
.SkipWhile(x => !x.IsPalindrome()) | ||
.Any()); | ||
} | ||
} | ||
} |
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