ProgrammingLogic문제 [yong27]의 [Python]버젼

1) 가장기본적인...
{{{
@@@@@@@
@@@@@@
@@@@@
@@@@
@@@
@@
@
}}}

{{{#!python
def ProOne():
    for i in range(7,0,-1):
        print '@'*i
}}}

2) 인덱스를 두개(i,j)를 돌린다는게 좀 맘에걸림
{{{
@@@@@@@
 @@@@@
  @@@
   @
}}}

{{{#!python
def ProTwo():
    j=0
    for i in range(7,0,-1):
        if i%2 == 1:
            print ' '*j+'@'*i
            j+=1
}}}
----
{{{
>>> for linei in range(4):
	stars=7-linei*2
	print ' '*linei+'*'*stars
}}}
--[김창준]
----
3) 첫째줄을 같이 쓸수 있는 로직이 있을까 고민했으나 못찾음
{{{
@@@@@@@
@@
@ @
@  @
@   @ 
@    @
@     @
}}}

{{{#!python
def ProThree():
    for i in range(7):
        if i==0:
            print '@'*7
        else: 
            print '@'+' '*(i-1)+'@'
}}}
----
{{{
>>> for i in range((7+1)*7):
	row,col=divmod(i,8)
	if col>=7: c='\n'
	else:
		c= (not row or not col or row==col) and '*' or ' '
	sys.stdout.write(c)
}}}
--[김창준]
----
and then now a more general approach is possible:

{{{
def printCells(aRow,aCol,aIsStar):
	for i in range((aCol+1)*aRow): 
		row,col=divmod(i,aCol+1)
		if col>=aCol: c='\n' 
		else: 
			c= aIsStar(row,col) and '*' or ' ' 
		sys.stdout.write(c)


>>> printCells(7,7,lambda row,col:-col+6>=row)
>>> printCells(4,7,lambda row,col:((col-3)+3 >= row and (3-col)+3 >= row ))
>>> printCells(7,7,lambda row,col:not row or not col or row==col)
}}}

I would call this a paradigm(esp. a Cartesian) of programming, which is a way of solving a category of similar problems, as in Seminar:TheParadigmsOfProgramming.

--[김창준]