Given an odd value of N, the program must print multi
layered rhombus pattern in diamond shapes whose side contains N, N-2, ... 1
slashes respectively as shown below in the examples.
Input
Format:
The first line contains N.
The first line contains N.
Output
Format:
The multi layered rhombus pattern in diamond shapes whose side contains N, N-2, ... 1 slashes respectively. Hash symbol is used as filler for other values.
The multi layered rhombus pattern in diamond shapes whose side contains N, N-2, ... 1 slashes respectively. Hash symbol is used as filler for other values.
Boundary Conditions:
1 <= N <= 101 and N is odd.
1 <= N <= 101 and N is odd.
Example
Input/Output:
Input:
5
Input:
5
Output:
####/\####
###/##\###
##/#/\#\##
#/#/##\#\#
/#/#/\#\#\
\#\#\/#/#/
#\#\##/#/#
##\#\/#/##
###\##/###
####\/####
####/\####
###/##\###
##/#/\#\##
#/#/##\#\#
/#/#/\#\#\
\#\#\/#/#/
#\#\##/#/#
##\#\/#/##
###\##/###
####\/####
Solution:
n
= int(input())
for
i in range(2*n):
for j in range(2*n):
if i<n and j<n and j>=n-i-1
and (j-n+1+i)%2==0:
print('/',end='')
elif i<n and j>=n and j<=n+i
and (j-n-i)%2==0:
print('\\',end='')
elif i>=n and j<n and j>=i-n
and (j-i+n)%2==0:
print('\\',end='')
elif i>=n and j>=n and
j<=3*n-i-1 and (3*n-i-1-j)%2==0:
print('/',end='')
else:
print('#',end='')
print()