【发布时间】:2015-11-27 04:04:55
【问题描述】:
尝试自学 ruby - 我正在研究 ruby 中的 Project Euler 问题 14。
n = 1000000
array = Array.new(n,0)
#array[x] will store the number of steps to get to one if a solution has been found and 0 otherwise. x will equal the starting number. array[0] will be nonsensical for these purposes
i = n-1#We will start at array[n-1] and work down to 1
while i > 1
if array[i] == 0
numstep = 0 #numstep will hold the number of loops that j makes until it gets to 1 or a number that has already been solved
j = i
while j > 1 && (array[j] == 0 || array[j] == nil)
case j%2
when 1 # j is odd
j = 3*j + 1
when 0 # j is even
j = j/2
end
numstep += 1
end
stop = array[j] #if j has been solved, array[j] is the number of steps to j = 1. If j = 1, array[j] = 0
j = i
counter = 0
while j > 1 && (array[j] == 0 || array[j] == nil)
if j < n
array[j] = numstep + stop - counter #numstep + stop should equal the solution to the ith number, to get the jth number we subtract counter
end
case j%2
when 1 #j is odd
j = 3*j+1
when 0 #j is even
j = j/2
end
counter += 1
end
end
i = i-1
end
puts("The longest Collatz sequence starting below #{n} starts at #{array.each_with_index.max[1]} and is #{array.max} numbers long")
此代码适用于 n = 100000 及以下,但当我上升到 n = 1000000 时,它会运行一段时间(直到 j = 999167 *3 + 1 = 2997502)。当它尝试访问数组的第 2997502 个索引时,它会抛出错误
in '[]': bignum too big to convert into 'long' (RangeError)
在第 27 行(这是 while 语句:
while j > 1 && (array[j] == 0 || array[j] == nil)
我怎样才能让它不引发错误?检查数组是否为零可以节省代码效率,因为它允许您不重新计算已经完成的事情,但是如果我删除 and 语句,它会运行并给出正确的答案。我很确定问题在于数组的索引不能是大数字,但也许有一种方法可以声明我的数组,这样它就可以了?我不太关心答案本身。我实际上已经在 C# 中解决了这个问题 - 只是想学习 ruby,所以我想知道我的代码为什么会这样做(如果我错了)以及如何解决它。
【问题讨论】:
-
2997502不是太大,无法转换为long,它甚至不是一个特别大的数组。不知何故,您正在创建一个更大的数字。 -
您的程序对我来说运行良好,并打印出正确的结果。您使用的是哪个版本的 ruby?