题解 | #2的幂次方#
2的幂次方
https://www.nowcoder.com/practice/7cf7b0706d7e4b439481f53e5fdac6e7
import math
list1 =['2(0)','2','2(2)','2(2+2(0))','2(2(2))','2(2(2)+2(0))','2(2(2)+2)','2(2(2)+2+2(0))','2(2(2+2(0)))','2(2(2+2(0))+2(0))','2(2(2+2(0))+2)','2(2(2+2(0))+2+2(0))','2(2(2+2(0))+2(2))','2(2(2+2(0))+2(2)+2(0))','2(2(2+2(0))+2(2)+2)']
n = eval(input())
list2 = []
ans = ''
inx = 0
for i in range (14,-1,-1):
if int(math.pow(2,i)) <= n:
list2.append(i)
n -= int(math.pow(2,i))
for i in range(len(list2)):
inx = list2[i]
ans += list1[inx]
if i != len(list2) - 1:
ans += '+'
print(ans)