4.11 网易互娱2020届校招补录笔试-服务端笔经-

#
# 接收两个表示9进制数的字符串，返回表示它们相加后的9进制数的字符串
# @param num1 string字符串 第一个加数
# @param num2 string字符串 第二个加数
# @return string字符串
#
class Solution:
    def add(self , num1 , num2 ):
        p1 = num1.find('.')
        p2 = num2.find('.')
        if p1!=-1 or p2!=-1:
            if p1==-1 and p2!=-1:
                num1,num2=num1,num2
            if p1!=-1 and p2==-1:
                int_num1, par_num1 = num1.split('.')
                return self.int_add(int_num1,num2)+'.'+par_num1
            if p1!=-1 and p2!=-1:
                int_num1, par_num1 = num1.split('.')
                int_num2, par_num2 = num2.split('.')
                l1 = len(par_num1)
                l2 = len(par_num2)
                l = max(l1,l2)
                par_num1+='0'*(l-l1)
                par_num2+='0'*(l-l2)
                sa = self.int_add(par_num1,par_num2)
                sb = self.int_add(int_num1,int_num2)
                if len(sa)>l:
                    
                    return self.int_add(sb,"1") + '.' + sa[1:]
                else:
                    return sb+'.'+sa
        else:
            return self.int_add(num1,num2)
    
    def int_add(self, num1, num2):
        n1 = len(num1)
        n2 = len(num2)
        s1 = n1 - 1
        s2 = n2 - 1
        res = []
        car = 0
        while s1>=0 or s2>=0 or car>0:
            if s1>=0:
                car += int(num1[s1])
                s1-=1
            if s2>=0:
                car += int(num2[s2])
                s2-=1
            d = car%9
            res.append(str(d))
            car//=9
        return "".join(reversed(res))
        

第二题

import bisect
n = int(input())
W = list(map(lambda x:int(x), input().split()))
T = list(map(lambda x:int(x), input().split()))
M = int(input())
W.sort()
T.sort()

def main():
    dp = 1
    for i in range(n):
        if W[i]<T[i]:
            return 0
        # 和 c++ std::lower_bound 一样, 二分查找第一个大于等于目标值的元素的索引
        can = i - bisect.bisect_left(W,T[i]) + 1
        dp = dp*can%M
    return dp

print(main())

第三题

ABCD
efgh
IJKL

n,m = map(lambda x:int(x), input().split())

job_player = [None]*n
for i in range(n):
    job_player[i] = list(input())

ruleY=[]
ruleN=[]
# print(job_player)
while True:
    
    i,j,c,p,q = input().split()
    if int(i)==0:
        break

    if c=='Y':
        ruleY.append(tuple(map(lambda x:int(x)-1,(i,j,p,q))))
    else:
        ruleN.append(tuple(map(lambda x:int(x)-1,(i,j,p,q))))

from itertools import permutations
from itertools import product

def judge(base):
    for i,j,p,q in ruleY:
        if base[i].index(j)!=base[p].index(q):
            return False
    for i,j,p,q in ruleN:
        if base[i].index(j)==base[p].index(q):
            return False
    return True

def myprint(base):
    for i in range(m):
        for j in range(n):
            print(job_player[j][base[j][i]],end='')
        print()

def brute_force():
    for base in product(permutations(range(m)),repeat=n):
        if judge(base):
            myprint(base)
            break

brute_force()

第四题

能100%AC的代码

class Solution:
    def maxBoxes(self , boxes ):
        boxes.sort()
        n=len(boxes)
        def can_be_contain(box1, box2):
            return all(box1[i]<box2[i] for i in range(3))

        dp = [1]*(1+n) # dp[i]表示以索引为i-1的盒子作为最外层盒子, 最多能打包多少盒子
        dp[0]=0
        for i in range(1, n+1):
            for j in range(i):
                if can_be_contain(boxes[j-1], boxes[i-1]):
                    dp[i] = max(dp[i], 1+dp[j])
        return max(dp) # 我考的时候return了dp[n-1]