k=int(input('value of a: '))
for a in range(1,k):
for b in range(1,k):
if a==b:
print('A',end=" ")
else:
print('B',end=" ")
print()
________________________________________________________________________________rows = 6 #Pattern(1,121,12321,1234321,123454321)
for i in range(1, rows + 1):
for j in range(1, i - 1):
print(j, end=" ")
for j in range(i - 1, 0, -1):
print(j, end=" ")
print()j="ramkumar"
for i in range(len(j)):
for g in range(0,i+1):
print(j[g],end=' ')
print('')
# end='' meaning (print the next answer , near the previous answer in same line)
#answer 1 2 3 4 5 6 like this in same row or same line(in horizontal)
name="ramkumar"
for i in range(len(name)) :
print(name[0:i+1])def pattern(n):
for i in range(0,n):
for j in range(0, i+1):
print("* " , end="")
print("\r") pattern(5
numbers = 10
for rowIndex in range(numbers):
rowString = ''
for columnIndex in range(numbers): #Same number of points
#each row is equivalent to the previous row + 1 - use existing iterator
rowString += str(columnIndex + rowIndex) + '\t'
print(rowString)
python
import re
v = "aeiou"
c = "qwrtypsdfghjklzxcvbnm"
print(*re.findall("(?=[%s]([%s]{2,})[%s])"%(c,v,c),input(), re.I) or [-1], sep="\n")