def maximumProduct(self, nums: list) -> int:
# 392ms 33.44%
# 找到最大的3个和最小的四个数后暴力枚举
from itertools import permutations as pr
max_three, min_four = [], []
target = []
if len(nums) > 7:
for num in nums:
max_three.append(num)
min_four.append(num)
max_three.sort(reverse=True)
min_four.sort()
if len(max_three) > 3:
max_three.pop()
if len(min_four) > 4:
min_four.pop()
target.extend(max_three)
target.extend(min_four)
else:
target = nums
print(target)
ans_max = target[0] * target[1] * target[2]
for couple in pr(target, 3):
# print(couple)
temp = couple[0]*couple[1]*couple[2]
ans_max = max(temp, ans_max)
return ans_max
def length(digits):
outputs = set()
ops = [o.add, o.sub, o.mul, o.truediv]
for a, b, c, d in p(digits):
for op1, op2, op3 in pr(ops, repeat=3):
outputs.update(
[op1(op2(op3(a, b), c), d),
op1(op2(a, b), op3(c, d))])
return next(i for i in co(1) if i not in outputs) - 1
def largestTimeFromDigits(self, A: List[int]) -> str:
x = ':'.join(
reduce(
lambda a, b: max([a, b]),
filter(
lambda t: t[0] <= '23' and t[1] <= '59',
map(
lambda x:
(str(x[0]) + str(x[1]), str(x[2]) + str(x[3])),
list(pr(A)))), ('', '')))
return x if x != ':' else ''
def __init__(self, symbolic_system):
self.A = {}
self.B_u = {}
self.B_lambda = {}
self.c = {}
self.Eta = []
self.E = {}
self.e = {}
self.symbolic_system = symbolic_system
self.list_of_contact_points = symbolic_system.list_of_contact_points
self.n, self.m_u, self.m_lambda = symbolic_system.n, symbolic_system.m_u, symbolic_system.m_lambda
self.Sigma = list(pr(*[i.Sigma for i in self.list_of_contact_points]))
def process():
try:
l = list(st.get())
n = nu.get()
from itertools import permutations as pr
tot = pr(l, n)
with open('outw.txt', 'w+') as f, open("dict-a_to_z.txt", 'r') as dic:
f.truncate()
dici = dic.read().strip().split()
for i in tot:
s = ""
for j in (i):
s = s + j
if str(s) in dici:
f.write(str(s) + "\n")
f.seek(0)
te = f.read()
f.close()
E3.insert(tkinter.END, te)
except:
messagebox.showwarning("Type correct details!!!")
from itertools import combinations_with_replacement as H
from itertools import product as pr
BB = ['Ab', 'Bb', 'Cb', 'Db']
WW = ['Aw', 'Bw', 'Cw', 'Dw']
BBbc = list(H(BB, 6))
WWwc = list(H(WW, 6))
total = list(pr(BBbc, WWwc))
WB = []
for i in total:
WB.append(i[0] + i[1])
WBA = [] # WBA=[x for x in WB if len(set([y[0] for y in x]))==4]
for x in WB:
if len(set([
y[0] for y in x
])) == 4 and x.count('Ab') >= 4 and x.count('Ab') > x.count('Aw'):
WBA.append(x)
WBAB = []
for i in WBA:
if ((i.count('Db') > i.count('Dw')) + (i.count('Cb') > i.count('Cw')) +
(i.count('Bb') > i.count('Bw')) == 1):
WBAB.append(i)
print(len(WBAB))
apply_const = lambda consts, const: [c for c in consts if c.isdisjoint(const)]
find_const = lambda key, consts: [c for c in consts if key in c]
def solve(sol, cs):
if len(cs) == 0: return sol if len(sol)==81 else None
cnt = Counter()
for c in cs: # Find best split
for k in c: cnt[k] += 1
for c in find_const(cnt.most_common()[-1][0], cs):
s = solve(sol+[c], apply_const(cs, c))
if s: return s
with open('p96.txt') as fin: sudoku = [line.strip() for line in fin]
consts = [make_const(i,j,n) for i,j,n in pr(range(9), range(9), range(1,10))]
total = 0
for start in range(1, len(sudoku), 10):
initial = []
for j,i in pr(range(9), range(9)):
n = int(sudoku[start+j][i])
if n: initial.append(make_const(i,j,n))
s = {}
for c in solve(initial, reduce(apply_const, initial, consts.copy())):
parts = sorted(c)
s[(parts[0][1], parts[0][2])] = parts[1][2]
total += s[(0,0)]*100 + s[(1,0)]*10 + s[(2,0)]
print(total)
def findSolution(self, customfunction, z):
return [
[i, j]
for i, j in pr(range(1, z + 1), repeat=2)
if customfunction.f(i, j) == z
]
'''
흰 공 개와 검은 공 개를 세 상자 A , B, C에 남김없이
나누어 넣을 때, 각 상자에 공이 개 이상씩 들어가도록 나누어
넣는 경우의 수를 구하시오. (단, 같은 색 공끼리는 서로 구별하지
않는다.)
개수가 적은 흰색부터 배열
4, 0, 0: 3 * (2군데에 2개 검은공)3H2 = 18
2, 2, 0: 3 * 3H4 = 45
1, 3, 0: 3! * 3H3 = 60
2, 1, 1: 3 * 3H4 = 45
=> 168가지
'''
from itertools import combinations_with_replacement as H
from itertools import product as pr
W=['Aw','Bw','Cw']
B=['Ab','Bb','Cb']
WW=list(H(W,4))
BB=list(H(B,6))
WB=list(pr(WW,BB))
totalWB=[i[0]+i[1] for i in WB]
WB2=[]
for i in totalWB:
a=[x[0] for x in i]
if a.count('A')>=2 and a.count('B')>=2 and a.count('C')>=2:
WB2.append(i)
print(len(WB2))
ciph = [int(a) for a in open("p059_cipher.txt").read().split(',')]
shortciph = ciph[:20]
# print(ciph[:20])
from itertools import product as pr
from string import ascii_lowercase
keys = [a for a in pr(ascii_lowercase,repeat=3)]
words = ['the', 'of', 'to', 'and', 'in']
# for a in keys:
# longkey = a*7
# plain = ''.join([chr(ord(k)^c) for k,c in zip(longkey, shortciph)])
# if any([w in plain for w in words]):
# print(a, plain)
longkey = "god"*(1+len(ciph)/3)
plain = ''.join([chr(ord(k)^c) for k,c in zip(longkey, ciph)])
print(plain)
print(sum([ord(c) for c in plain]))
data2 = [] # . 개수
data3 = [] # 가능한 R,C,S
for _ in 'a' * p:
s = input()
s2 = s.lstrip('.')
dot = len(s) - len(s2)
for i in s:
if i == '(': a += 1
if i == ')': a -= 1
if i == '{': b += 1
if i == '}': b -= 1
if i == '[': c += 1
if i == ']': c -= 1
data1.append([a, b, c])
data2.append(dot)
for j in pr(range(1, 21), repeat=3):
x, y, z = j # 가능한 R,C,S
for i in range(1, p):
if x * data1[i][0] + y * data1[i][1] + z * data1[i][2] != data2[i]:
break
else:
data3.append([x, y, z])
R = set()
C = set()
S = set()
for i in data3:
R.add(i[0])
C.add(i[1])
S.add(i[2])
if len(R) == 1:
R = R.pop()
import itertools
from itertools import product as pr
pete_poss = pr('1234', repeat=9)
colin_poss = pr('123456', repeat=6)
pete_sums = [0] * 28
colin_sums = [0] * 31
for i in pete_poss:
pete_sums[sum([int(j) for j in list(i)]) - 9] += 1
for i in colin_poss:
colin_sums[sum([int(j) for j in list(i)]) - 6] += 1
P_pete = 0
runsum = sum(colin_sums[:3])
for n in range(9, 37):
P_pete += (pete_sums[n - 9] * runsum)
runsum += colin_sums[n - 6]
'''
P_tie = 0
for n in range(9,37):
P_tie += (pete_sums[n - 9] * colin_sums[n - 6])
P_colin = 0
runsum = colin_sums[30]
for n in range(35, 8, -1):
P_colin += (pete_sums[n - 9] * runsum)
runsum += colin_sums[n - 6]
'''
denom = 262144 * 46656
from itertools import permutations as pr perm = pr([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) arr = list(perm) length = len(arr) print arr[999999]
def permutations(s):
return [''.join(i) for i in sorted(set(pr(s)))]
#!/usr/bin/env python
# coding: utf-8
# In[28]:
a,b=input().split(' ')
i=int(input())
from itertools import product as pr
comb=pr(list(range(int(a),int(b)+1)),repeat=i)
count=0
for j in comb:
if sum(j)%2==0:
count=count+1
print(count%1000000007)
from itertools import combinations_with_replacement as H #중복조합
from itertools import product as pr
BB = ['Ab', 'Bb', 'Cb', 'Db'] #'Ab'는 A가 검은모자 갖기
WW = ['Aw', 'Bw', 'Cw', 'Dw']
totalB = list(H(BB, 6)) #검은색모자나눠주기
totalW = list(H(WW, 6)) #흰색모자나눠주기
totalBW = list(pr(totalB, totalW))
BW = [i[0] + i[1] for i in totalBW] #튜플을 합쳐준다 조건제시법
BWGa = []
for i in BW:
if set([x[0] for x in i
]) == {'A', 'B', 'C', 'D'} and i.count('Ab') >= 4 and i.count(
'Ab') > i.count('Aw'): # len(set([x[0] for x in i]))==4
BWGa.append(i)
BWGaB = []
for i in BWGa:
if i.count('Bb') > i.count('Bw') and i.count('Cb') <= i.count(
'Cw') and i.count('Db') <= i.count('Dw'):
BWGaB.append(i)
BWGaC = []
for i in BWGa:
if i.count('Cb') > i.count('Cw') and i.count('Bb') <= i.count(
'Bw') and i.count('Db') <= i.count('Dw'):
BWGaC.append(i)
BWGaD = []
for i in BWGa:
if i.count('Db') > i.count('Dw') and i.count('Cb') <= i.count(
'Cw') and i.count('Bb') <= i.count('Bw'):
BWGaD.append(i)
len(BWGaB + BWGaC + BWGaD)
s = list(input())
from itertools import product as pr
p = list(pr(['+', '-'], repeat=3))
for ope in p:
f = s[0] + ope[0] + s[1] + ope[1] + s[2] + ope[2] + s[3]
if eval(f) == 7:
print(f + '=7')
exit(0)
