[Python知识库]腾讯技术研究类和数据分析第一次笔试（2021.8.22）——Python

第一题：开锁——数学期望

# 最优策略：钥匙的选择先从消耗时间最少的开始选择，然后选择第二小的依次类推
# 开锁概率1/n
def openLockTime(n, m, time):
	time_reverse = []  # (n,m)->(m,n)
	for i in range(m):
		m_time = []
		for j in range(n):
			m_time.append(time[j][i])
		time_reverse.append(sorted(m_time))
	E = 0.000000
	p = 1 / n
	for i in range(m):
		i_sum = 0
		for j in range(n):
			i_sum = i_sum + sum(time_reverse[i][:j + 1])
		E += p * i_sum
	return E


# 钥匙数、锁个数
n, m = list(map(int, input().split()))
# 第i把钥匙打开第j把锁的时间
time = []
for i in range(n):
	t = list(map(int, input().split()))
	time.append(t)
print(openLockTime(n,m,time))
'''
2 4
347 177 40 84
107 282 347 193
'''

# 对time进行翻转


# 钥匙数、锁个数
n, m = list(map(int, input().split()))
# 第i把钥匙打开第j把锁的时间
time = []
for i in range(n):
	t = list(map(int, input().split()))
	time.append(t)

第二题：马夫**链——条件概率（矩阵相乘）

def predictOwnP(step, last, P):
	# 第i-1步last到第i步cur的状态
	for i in range(step):
		stab = 0.0
		Own = 0.0
		Fail = 0.0
		for j in range(3):
			stab += P[0][j] * last[j]
			Fail += P[1][j] * last[j]
			Own += P[2][j] * last[j]
		last = [stab, Fail, Own]
	if last[-1] > 0.5:
		return 1
	else:
		return 0

group_num = int(input())
for i in range(group_num):
	step = int(input())
	first = list(map(float, input().split()))
	P = []
	for _ in range(3):
		P.append(list(map(float, input().split())))
	print(predictOwnP(step, first, P))

'''
2
2
0.4 0.2 0.4
0.2 0.1 0.7
0.3 0.5 0.2
0.5 0.4 0.1
10
0.4 0.2 0.4
0.2 0.1 0.7
0.3 0.5 0.2
0.5 0.4 0.1
'''

第三题：迎宾车队——极差（双指针）

def carMaxNum(n,speed):
	speed = sorted(speed)
	maxNum = 0
	left = 0
	right = 0
	while right < n:
		if speed[right] - speed[left] <= 10:
			maxNum = max(maxNum,right - left + 1)
			right += 1
		else:
			left += 1
	return maxNum

n = int(input())
speed = list(map(int,input().split()))
print(carMaxNum(n,speed))
'''
6
17 3 21 8 44 24
'''
'''
3
9 7 4
'''

第四题：水站的水流量——直接找的规律

def calcuMaxNode(n, t):
	# 第i层和第i-1层的水流支路有2i-2条，每条能分到1/(2i-2)个水流量，中间的结点每秒会得到1/(i-1)个水流量，因此需要i-1s才能满载
	# 外面的两个则需要2i-2s才能满载
	# 每一层满载需要消耗的时间为is
	maxT = sum([i+1 for i in range(n)])
	if t >= maxT:
		return maxT
	# 先确定目前流到了哪一层
	time = [1 for _ in range(11)]
	for i in range(2,11):
		time[i] = time[i-1] + i
	print(time)
	layers = 1
	for layer in range(1,n+1):
		if time[layer] > t:
			layers = layer # 流到了第layer层,且还没有流满
			break
	inside_time = layers/(layers - 1) # 里面水结点流满需要的时间
	outside_time = 2*layers/(layers - 1)  # 外面2个水结点流流满的时间
	if time[layers-1] + inside_time*(layers-2) > t:
		return sum([i for i in range(1,layers)])
	elif time[layers-1] + inside_time*(layers-2) <= t and time[layers] + outside_time*2 > t:
		return sum([i for i in range(1,layers+1)]) - 2
	else:
		return sum([i for i in range(1,layers+1)])

n, t = list(map(int, input().split()))
print(calcuMaxNode(n,t))

'''
3 5
'''

第五题：轰炸——DFS或者BFS（将1包围的0改为2）

def hongzha(matrix):
	m = len(matrix)
	n = len(matrix[0])
	# 先将没有被1包围的0给去掉，没有被1包围也就是可以通到矩阵的四边
	# 判断是否需要继续遍历
	def judge(indexi,indexj,hashtable,matrix):
		if indexi < 0 or indexj < 0 or indexi >= m or indexj >= n:
			return False
		if hashtable[indexi][indexj] == True:
			return False
		if matrix[indexi][indexj] == 1:
			return False
		return True

	# 广度优先搜索
	def bfs(indexi,indexj,hashtable,res):
		'''
		:param indexi:
		:param indexj:
		:param hashtable:
		:param P: 存放着四边上的0元素坐标，后面会将不被包围的坐标也添加上去
		:return:
		'''
		P= [(indexi,indexj)]
		hashtable[indexi][indexj] = True
		I = [0,-1,1,0]
		J = [-1,0,0,1]
		while P:
			indexi,indexj = P[0]
			res.append(P[0])
			for i in range(4):
				if judge(indexi+I[i],indexj+J[i],hashtable,matrix):
					P.append((indexi+I[i],indexj+J[i]))
					hashtable[indexi+I[i]][indexj+J[i]] = True
			P = P[1:]

	hashtable = [[False for _ in range(n)] for _ in range(m)]
	res = []
	for i in range(m):
		if judge(i,0,hashtable,matrix):
			bfs(i,0,hashtable,res)
		if judge(i,n-1,hashtable,matrix):
			bfs(i, n-1, hashtable, res)
	for j in range(n):
		if judge(0,j,hashtable,matrix):
			bfs(0,j,hashtable,res)
		if judge(m-1,j,hashtable,matrix):
			bfs(m-1,j, hashtable, res)

	for i in range(m):
		for j in range(n):
			if matrix[i][j] == 0 and (i,j) not in res:
				matrix[i][j] = 2
	return matrix

import random
m = random.randint(5,11)
n = random.randint(5,11)
matrix = []
for i in range(m):
	line = []
	for j in range(n):
		line.append(random.randint(0,1))
	matrix.append(line)

newMatrix = hongzha(matrix)
for line in newMatrix:
	print(line)