字节跳动笔试08.28
第一题：20分，判断特征，哪些可以被计算，过了90%
n = int(input())
needs = [int(i) for i in input().split()]
relations = []
for i in range(n-1):
relations.append([int(i) for i in input().split()])
dic = {}
for relation in relations:
dic[relation[0]] = relation[1:]
ans = []
for j in needs:
flag = False
if j not in dic:
ans.append(1)
else:
seta = dic[j]
for item in seta:
if item in dic:
flag = True
setb = dic[item]
if j in setb:
ans.append(0)
else:
ans.append(1)
if not flag:
ans.append(1)
print(" ".join(str(i) for i in ans)) 第二题：25分，输出最大子数组乘积位置，过了50% n = int(input())
lens = []
subsets = []
for i in range(n):
len_num = int(input())
lens.append(len_num)
subsets.append([int(i) for i in input().split()])
def minmul(lis):
n = len(lis)
dp = [1]*n
l,r = 0,0
res,ans = [],[]
for i in range(n):
if lis[i]==0:
dp[i] = 0
else:
if i>=1 and lis[i-1]==0:
dp[i]= lis[i]
else:
dp[i]= dp[i-1]*lis[i]
res = max(dp)
for i,item in enumerate(dp):
if item==0:
l = i+1
if res==item:
r = i
ans.append([l+1,r+1])
return min(ans)
for subset in subsets:
print(" ".join(str(i) for i in minmul(subset))) 第三题，25分，翻转一次，输出最大子数组和 尝试暴力，过了20%
n = int(input()) nums = ([int(i) for i in input().split()]) l,r = 0,0 ans = 0 def maxsum(nums): # 子数组最大和 n = len(nums) dp = [0]*n for i in range(n): dp[i] = nums[i] for i in range(1,n): dp[i] += max(dp[i-1],0) return max(dp) for i in range(n): if nums[i]<0: l=i # 第二种是和后面的交换 for j in range(i+1,n): if nums[j]>0: r = j if l!=r: nums[l:r+1] = nums[l:r+1][::-1] ans = max(ans,maxsum(nums)) # 只能翻转一次，要翻转回去 nums[l:r+1] = nums[l:r+1][::-1] # 第一种是和前面的交换 for j in range(0,i): if nums[j]>0: r = j if l!=r: nums[l:r+1] = nums[l:r+1][::-1] ans = max(ans,maxsum(nums)) nums[l:r+1] = nums[l:r+1][::-1] print(ans)
第四题，30分，删去一个元素，输出K个元素子数组的最大和，暴力过了10%
