所有那些繁忙的海狸都弄得一团糟。他们在磁带上写满了。以这种速度，我们的邻居将停止向我们提供无限制的磁带。

我们需要一种新的方式来玩忙碌的海狸游戏，这种方式不会破坏我们使用的所有磁带。

规则

只需动脑子。内存磁带无界。输入指令将始终读取，因此可用于清除值。$0$

50个字节的源限制。

执行结束时，内存必须全为 s。$0$

分数是内存指针的起始位置和最终位置之间的距离-如果在它们之间需要移动指令，则分数为n。越高越好。如果可以，请提供准确的值，否则请提供估计值。$n$$n$

例

32个字节，$2^{255}-1$

-[-[[>]+>+[<]>-[[>]<+<+[<]>-]]>]

说明

-                                Initialize the list to [255].
 [                             ] Repeat as long as the list is not empty.
 [-                            ] Decrement the left end. We need to shrink the numbers so it ends eventually.
 [ [                         ] ] Skip if 0 already.
 [ [[>]                      ] ] Move to the cell past the right end.
 [ [   +                     ] ] Make this cell 1.
 [ [    >                    ] ] Go right again.
 [ [     +                   ] ] Make this cell 1. We've now appended [1, 1].
 [ [      [<]>               ] ] Go back to the first nonzero cell on the left.
 [ [          -              ] ] And decrement it.
 [ [           [            ]] ] We will need to transfer the rest of the number from the left to the right, so keep looping.
 [ [           [[>]<        ]] ] Go to the last nonzero cell on the right.
 [ [           [    +<+     ]] ] Increment this and the one on the left. These are the cells we appended earlier. We transfer to them.
 [ [           [       [<]> ]] ] Go back to the first nonzero cell on the left, which we are transferring from.
 [ [           [           -]] ] Decrement here on the left to balance out the incrementing on the right.
 [                            >] We end the iteration on a now empty cell. Move right, the new left end is there.

$[255]$$n$$n>1$$[n-1, n-1]$$(n-1)$$(n)$$1$$0$$[n]$$2^n-1$

本示例旨在说明创建提交时使用的一些技术。它的大小没有竞争力。

$A(255, 2) - 1 = (2 \uparrow^{253} 5) - 4$

+<+<<++>-[-[<+>>[-<[->+<]+>>]+<-]>[>]+[<]+<]>[->]

$A$[-[<+>>[-<[->+<]+>>]+<-]>[>]+[<]+<]$A(m,n)$1 - m, m, 1 <n times>m$A(m,n)$

我使用以下Python程序对程序的行为进行了建模：

def a(M, N):
    assert M > 0
    m = [-M + 1, M]
    n = N
    while m[-1]:
        while m[-1] > 1:
            m[-1] -= 1
            m[-2] += 1
            while n:
                m.insert(-1, 1)
                n -= 1
            n = 1
        n += 2
        m.pop()
    return n