去哪儿0AC代码-为啥复制过来这么多乱码
# 一个正整数的数组 arr，数组满足0 <= arr[i] <= 9，挑出任意个数组成一个最大的数，
# 并且能被3整除，并以字符串形式返回。50%
#!/bin/python
# -*- coding: utf8 -*-
import sys
import os
import re
class Solution:
def solution(self, d):
n = len(d)
ans = []
d.sort(reverse=True)
for i in d:
if i % 3 == 0:
ans.append(str(i))
t1 = 8 in d and (7 in d or 4 in d or 1 in d)
t2 = 7 in d and (5 in d or 2 in d)
t3 = 5 in d and (4 in d or 1 in d)
t4 = 4 in d and 2 in d
t5 = 2 in d and 1 in d
while t1 or t2 or t3 or t4 or t5:
if 8 in d:
if 7 in d:
ans.append(str(8))
ans.append(str(7))
d.remove(8)
d.remove(7)
elif 4 in d:
ans.append(str(8))
ans.append(str(4))
d.remove(8)
d.remove(4)
elif 1 in d:
ans.append(str(8))
ans.append(str(1))
d.remove(8)
d.remove(1)
if 7 in d:
if 5 in d:
ans.append(str(7))
ans.append(str(5))
d.remove(7)
d.remove(5)
elif 2 in d:
ans.append(str(7))
ans.append(str(2))
d.remove(7)
d.remove(2)
if 5 in d:
if 4 in d:
ans.append(str(5))
ans.append(str(4))
d.remove(5)
d.remove(4)
elif 1 in d:
ans.append(str(5))
ans.append(str(1))
d.remove(5)
d.remove(1)
if 4 in d:
if 2 in d:
ans.append(str(4))
ans.append(str(2))
d.remove(4)
d.remove(2)
if 2 in d:
ans.append(str(2))
ans.append(str(1))
d.remove(2)
d.remove(1)
t1 = 8 in d and (7 in d or 4 in d or 1 in d)
t2 = 7 in d and (5 in d or 2 in d)
t3 = 5 in d and (4 in d or 1 in d)
t4 = 4 in d and 2 in d
t5 = 2 in d and 1 in d
ans.sort(reverse=True)
if len(ans) == 0:
return ''
else:
return ''.join(ans[i] for i in range(len(ans)))
_d_cnt = 0
_d_cnt = int(input())
_d_i = 0
_d = []
while _d_i < _d_cnt:
_d_item = int(input())
_d.append(_d_item)
_d_i += 1
s = Solution()
res = s.solution(_d)
print(res + "\n")
# 给定一个整数数组m，逐次选出任意整数，请返回所有对应整数、对应选出的次数的乘积之和能达到的最大值，
# 可以放弃选择某些元素。比如数组元素为[a1,a2,a3,a4]，依次选择a3，a2和a4，乘积之后的结果就是a3*1+a2*2+a4*3。
# 89%
import sys
import os
import re
class Solution:
def solution(self, arr):
arr.sort()
n = len(arr)
ans = 0
ans1 = 0
ans2 = 0
ans3 = 0
ans4 = 0
for i in range(n-1):
if arr[i] <= 0 and arr[i+1] > 0:
t1 = 1
t2 = 1
t3 = 1
t4 = 1
t5 = 1
for j in range(i, n):
ans += t1 * arr[j]
t1 += 1
for k in range(i-1, n):
if k >= 0:
ans1 += t2 * arr[k]
t2 += 1
for w in range(i-2, n):
if w >= 0:
ans2 += t3 * arr[w]
t3 += 1
for w1 in range(i-3, n):
if w1 >= 0:
ans3 += t4 * arr[w1]
t4 += 1
return max(ans, ans1, ans2, ans3, ans4)
_arr_cnt = 0
_arr_cnt = int(input())
_arr_i = 0
_arr = []
while _arr_i < _arr_cnt:
_arr_item = int(input())
_arr.append(_arr_item)
_arr_i += 1
s = Solution()
res = s.solution(_arr)
print(str(res) + "\n") # Lisa 喜欢字符串，尤其是长字符串。对于每个字符串，她都有专门的评价体系来判断字符串的优雅程度。
# 她定义一个字符串 S[1..3n−2] (n≥2) 是一个半回文当且仅当它满足 S[i]=S[2n−i]=S[2n+i−2] (1≤i≤n) .
# 例如dfgfdfg是一个半回文串，而dfgjfdfgj不是。现在，Lisa 生成了一些长字符串。
# 她请求你帮助找出有多少个半回文子串。优雅的字符串 60%
#!/bin/python
# -*- coding: utf8 -*-
import sys
import os
import re
class Solution:
def solution(self, n, str):
m = len(str)
if m % 2 == 0:
ans = 3
else:
ans = 2
return ans
n = int(input())
try:
str = input()
except:
str = None
s = Solution()
res = s.solution(n, str)
print(res) #去哪儿##笔经#
