def specialCombinations(vec):
b = []
j = 0
for a in cr(vec, 4):
b.append(a)
j += 1
return b
def specialCombinations(self):
"""This function computes the number of possible configurations
of N basis functions into poducts of four functions."""
vec = list(xrange(self.N))
b = []
# for a1 in cr(vec, 2):
# for a2 in cr(vec, 2):
# b.append(a1+a2)
for a in cr(vec, 4):
b.append(a)
return b
def specialCombinations(self):
"""This function computes the number of possible configurations
of N basis functions into poducts of four functions."""
vec = list(xrange(self.N))
b = []
# for a1 in cr(vec, 2):
# for a2 in cr(vec, 2):
# b.append(a1+a2)
for a in cr(vec, 4):
b.append(a)
return b
def calculate(self):
loop = cr(range(len(self.X)),2)
L = 0
if np.linalg.norm(self.X-self.X_) == 0: # self inductance correction
L += (np.sum(np.linalg.norm(self.dX,axis=1)))/2
mutual = False
else:
mutual = True
for ij in loop:
if ij[0] != ij[1] or mutual:
L += self.intergrate(ij,self.X,self.X_,self.dX,self.dX_,self.r)
else:
dl = np.linalg.norm(self.dX[ij[0],:])
if dl > self.r:
L += 2*np.log(dl/self.r)
#L += self.substeps(ij) # self inductance sub-calc
L *= cc.mu_o/(4*np.pi)
return L
def main():
num_of_testcase = int(stdin.readline())
for _ in range(num_of_testcase):
number = int(stdin.readline())
prime_list = get_prime_list(number)
prime_comb = list(cr(prime_list, 2))
for comb in prime_comb:
if number - sum(comb) in prime_list:
third_number = number - sum(comb)
answer = ''
for num in comb:
answer += str(num) + ' '
print(answer.rstrip() + ' ' + str(third_number))
break
else:
print(0)
# project euler problem 23
# http://projecteuler.net/problem=23
# brute force, finished in 13.1s on my computer. Need to optimize
# Is it possible to avoid computing the abundant numbers in the first place ?
# NOTE: using generators instead of lists can save some CPU time.
from itertools import combinations_with_replacement as cr
END = 28123
def is_abundant(n):
s = set()
for i in xrange(1, int(n**0.5) + 1):
if n % i == 0:
s.add(i)
s.add(n/i)
s.remove(n)
return sum(s) > n
nums = [False] * END
abundant_numbers = (x for x in xrange(1, END) if is_abundant(x))
sieve = (sum(i) for i in cr(abundant_numbers, 2) if sum(i) < END)
for i in sieve:
nums[i] = True
print sum(i for i in xrange(1, END) if not nums[i])
def source():
data = input_data()
exc = data[1:]
points = list(i for i in range(1, data[0] + 1) if i not in exc)
return len(list(cr(points, 3)))
from itertools import combinations_with_replacement as cr
x, y = input().split()
x = sorted(x)
for i in cr(x, int(y)):
print("".join(i))
# problem-link = https://www.hackerrank.com/challenges/itertools-combinations-with-replacement/problem
from itertools import combinations_with_replacement as cr
input_value = list(input().strip().split())
for i in list(cr(sorted(input_value[0]), int(input_value[1]))):
print(''.join(i))
from itertools import combinations_with_replacement as cr
import string
alpa = list(zip(string.ascii_lowercase,list(range(1,27))))
s = ''
for t in range(int(input())):
n, k = map(int, input().split())
pr = [i for i in list(cr(range(1, 27), n)) if sum(i) == k][0]
for i in pr:
for a in alpa:
if i == a[1]:
s = s + a[0]
print(s)
s = ''
from itertools import combinations_with_replacement as cr
word, num = input().split()
print('\n'.join(''.join(i) for i in cr(sorted(word), int(num))))
Print the combinations with their replacements of string S on separate lines.
Sample Input:
HACK 2
Sample Output:
AA
AC
AH
AK
CC
CH
CK
HH
HK
KK
"""
from itertools import combinations_with_replacement as cr
S, num = list(input().split(' '))
string = []
for i in S:
string.append(i)
string.sort()
for j in list(cr(string, int(num))):
for k in j:
print(k, end='')
print('')
