Python实现小明种苹果算法：环形列表处理的简洁方案

for i in range(len(l)):
 if l[i-2] and l[i-1] and l[i]:
 e += 1

if N >= 3: # 苹果树少于三棵时 直接为0
 for y in range(len(is_fall)):
 if is_fall[y%len(is_fall)] == is_fall[(y+1)%len(is_fall)] == is_fall[(y+2)%len(is_fall)] == 1:
  E += 1
else:
 E = 0

res_drop.append(res_drop[0])
res_drop.append(res_drop[1])
for i in range(N):
 if res_drop[i] == 1 and res_drop[i+1] == 1 and res_drop[i+2] == 1:
 num += 1
print(' '.join([str(res),str(count_drop),str(num)]))

for i in range(len(is_fall)):
 pre=i-1
 if pre<0:
 pre+=len(is_fall)
 next=i+1
 if next > len(is_fall)-1:
 next-=len(is_fall)
 if is_fall[pre]==is_fall[i]==is_fall[next]==1:
 E+=1

n = int(input())
result = 0 # 总共剩余的苹果
count = 0
e = 0 # 连续三颗树的掉落情况
dl = 0 # 掉落的苹果棵树
l = [0] * n # n颗苹果树是否有苹果掉落的情况
for i in range(n):
 fs = []
 zs = []
 s = total = 0
 x = list(map(int,input().split()))
 x = x[::-1]
 caozuo = x.pop()
 #判断剩余的苹果个数
 for j in range(caozuo):
 s = x[j]
 if s <= 0:
  fs.append(s)
  #print(fs)
 else:
  zs.append(s)
  break

 result += sum(fs) + zs[0]

 total = x.pop()
 ss = 0
 for j in range(len(x)):
 flag = z = 0
 s = x.pop()
 if s < 0:
  ss = ss + s
 elif s == 0:
  continue
 else:
  z = s
  total = total + ss
  if total > z:
  flag = 1
  else:
  flag = 0
 l[i] = flag
 if flag == 1:
  dl += 1
  break
 else:
  continue
  
for i in range(len(l)):
 if l[i-2] and l[i-1] and l[i]:
 e += 1
print(result,dl,e)