第一題：開鎖——數(shù)學(xué)期望

# 最優(yōu)策略：鑰匙的選擇先從消耗時(shí)間最少的開始選擇，然后選擇第二小的依次類推 # 開鎖概率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 = 0.000000p = 1 / nfor i in range(m):i_sum = 0for j in range(n):i_sum = i_sum + sum(time_reverse[i][:j + 1])E += p * i_sumreturn E# 鑰匙數(shù)、鎖個(gè)數(shù) n, m = list(map(int, input().split())) # 第i把鑰匙打開第j把鎖的時(shí)間 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 '''# 對(duì)time進(jìn)行翻轉(zhuǎn)# 鑰匙數(shù)、鎖個(gè)數(shù) n, m = list(map(int, input().split())) # 第i把鑰匙打開第j把鎖的時(shí)間 time = [] for i in range(n):t = list(map(int, input().split()))time.append(t)

第二題：馬夫**鏈——條件概率（矩陣相乘）

def predictOwnP(step, last, P):# 第i-1步last到第i步cur的狀態(tài)for i in range(step):stab = 0.0Own = 0.0Fail = 0.0for 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 1else:return 0group_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 '''

第三題：迎賓車隊(duì)——極差（雙指針）

def carMaxNum(n,speed):speed = sorted(speed)maxNum = 0left = 0right = 0while right < n:if speed[right] - speed[left] <= 10:maxNum = max(maxNum,right - left + 1)right += 1else:left += 1return maxNumn = int(input()) speed = list(map(int,input().split())) print(carMaxNum(n,speed)) ''' 6 17 3 21 8 44 24 ''' ''' 3 9 7 4 '''

第四題：水站的水流量——直接找的規(guī)律

def calcuMaxNode(n, t):# 第i層和第i-1層的水流支路有2i-2條，每條能分到1/(2i-2)個(gè)水流量，中間的結(jié)點(diǎn)每秒會(huì)得到1/(i-1)個(gè)水流量，因此需要i-1s才能滿載# 外面的兩個(gè)則需要2i-2s才能滿載# 每一層滿載需要消耗的時(shí)間為ismaxT = sum([i+1 for i in range(n)])if t >= maxT:return maxT# 先確定目前流到了哪一層time = [1 for _ in range(11)]for i in range(2,11):time[i] = time[i-1] + iprint(time)layers = 1for layer in range(1,n+1):if time[layer] > t:layers = layer # 流到了第layer層,且還沒有流滿breakinside_time = layers/(layers - 1) # 里面水結(jié)點(diǎn)流滿需要的時(shí)間outside_time = 2*layers/(layers - 1) # 外面2個(gè)水結(jié)點(diǎn)流流滿的時(shí)間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)]) - 2else:return sum([i for i in range(1,layers+1)])n, t = list(map(int, input().split())) print(calcuMaxNode(n,t))''' 3 5 '''

第五題：轟炸——DFS或者BFS（將1包圍的0改為2）

def hongzha(matrix):m = len(matrix)n = len(matrix[0])# 先將沒有被1包圍的0給去掉，沒有被1包圍也就是可以通到矩陣的四邊# 判斷是否需要繼續(xù)遍歷def judge(indexi,indexj,hashtable,matrix):if indexi < 0 or indexj < 0 or indexi >= m or indexj >= n:return Falseif hashtable[indexi][indexj] == True:return Falseif matrix[indexi][indexj] == 1:return Falsereturn True# 廣度優(yōu)先搜索def bfs(indexi,indexj,hashtable,res):''':param indexi::param indexj::param hashtable::param P: 存放著四邊上的0元素坐標(biāo)，后面會(huì)將不被包圍的坐標(biāo)也添加上去:return:'''P= [(indexi,indexj)]hashtable[indexi][indexj] = TrueI = [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]] = TrueP = 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] = 2return matriximport 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)

總結(jié)

以上是生活随笔為你收集整理的腾讯技术研究类和数据分析第一次笔试（2021.8.22）——Python的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

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