题解 | #反复横跳#

整体思路，先构建出反复横跳的数组

[0, 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, -1365, 2731, -5461, 10923, -21845, 43691, -87381]

然后对把两个人的位置相减，可能是正也可能是负，假设这个数字为8200，则8200位于[2731,10923]之间，那应该取哪个呢？两种思路，要么多走一些，先走10923步，再走-2731步，要么少走一些，先走2731步，再走5469步，至于取哪种，由于时间紧迫，通过在数字比较小（3到11之间一个一个试）的时候观察大概推理得到如果一个数位于[2731,8192]之间，取2731，如果位于[8192,10923]之间，取10923，至于8192这个数字怎么来。8192=(10923-2731)/3*2+2731，即把[2731,10923]分成三份，前两份就取2731，后一份取10923。之后就不停迭代就好了，负数同理。

顺便建立一个字典用来处理[-5,5]的情况以提高效率。

x = 0
op = 1
step = 1
shuzu = [0]

for i in range(18):
    x += op * step
    op = -op
    step = step * 2
    shuzu.append(x)
x = list(map(int, input().split()))

zidian = {-5: 4,
          -4: 6,
          -3: 8,
          -2: 5,
          -1: 2,
          0: 0,
          1: 1,
          2: 3,
          3: 3,
          4: 5,
          5: 6}


cha = x[1] - x[0]
bushu = 0
flag = True

while flag:
    for i in range(1, len(shuzu)):
        if cha > 5:
            if shuzu[i - 1] <= cha < (shuzu[i + 1] - shuzu[i - 1]) / 3 * 2 + shuzu[i - 1]:
                bushu += i
                cha = cha - shuzu[i - 1]
                break
            if (shuzu[i + 1] - shuzu[i - 1]) / 3 * 2 + shuzu[i - 1] <= cha <= shuzu[i + 1]:
                bushu += i + 2
                cha = cha - shuzu[i + 1]
                break
        elif cha < -5:
            if shuzu[i - 1] >= cha >= (shuzu[i + 1] - shuzu[i - 1]) / 3 * 2 + shuzu[i - 1]:
                bushu += i
                cha = cha - shuzu[i - 1]
                break
            if (shuzu[i + 1] - shuzu[i - 1]) / 3 * 2 + shuzu[i - 1] > cha >= shuzu[i + 1]:
                bushu += i + 2
                cha = cha - shuzu[i + 1]
                break
        else:
            bushu += zidian[cha]
            flag = False
            break
print(bushu)