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 = 0.000000p = 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# 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 = 0.0Own = 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
'''

第三题：迎宾车队——极差（双指针）

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ismaxT = 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] + iprint(time)layers = 1for layer in range(1,n+1):if time[layer] > t:layers = layer # 流到了第layer层,且还没有流满breakinside_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)]) - 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 Falseif hashtable[indexi][indexj] == True:return Falseif matrix[indexi][indexj] == 1:return Falsereturn 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] = 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)