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maxlimit.py
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maxlimit.py
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import numpy as np
# huawei 面试题1
# 上游系统u's,共同调用核心服务,最大cnt个调用: sum(u's)<=cnt,不然给出limit,使得u_sorted[j:]=limit,求最大的limit
# 提示:limit有一个可变上界,由sum(u[0:i], (len-i)*limit) <= cnt决定,控制i,limit的最大值为最大上界
# 示例代码如下:
u = input("input nums: ")
u = [int(n) for n in u.split()]
cnt = int(input("count: "))
print(u, cnt)
length = len(u)
np.array(u).sort()
max = 0
for i in range(length):
candidate = (cnt - np.sum(u[:i])) / (length - i)
max = candidate if max < candidate else max
print(max)
# huawei 面试题2
# 输入:节点数、边数、连接关系、屏蔽节点数,屏蔽节点 输出:由根通向最短叶节点的路径
tree = {}
old_tree = {}
edges = []
node_num = int(input())
edge_num = int(input())
for i in range(edge_num):
edges.append(input().split())
block_num = int(input())
edges.sort(key=lambda x: int(x[0]))
for edge in edges:
flag = True
if edge[0] == '0':
tree["->".join(edge)] = 1
flag = False
else:
for key in list(tree.keys()):
if key[-1] == edge[0]:
cnt = tree.pop(key)
tree[key + "->" + edge[1]] = cnt + 1
old_tree[key] = cnt
flag = False
break
if flag:
for key in list(old_tree.keys()):
if key[-1] == edge[0]:
tree[key + "->" + edge[1]] = old_tree[key] + 1
break
for i in range(block_num):
blocked_node = input()
for key in list(tree.keys()):
if key.find(blocked_node) != -1:
tree.pop(key)
min, min_way = -1, "NULL"
for key in tree.keys():
if min == -1 or min > tree[key]: min, min_way = tree[key], key
print(min_way)