python咋实现输出*为(金字塔)形状,圣诞树形状。

模仿打印星星的例子,打印一个等腰三角形,(我们那个例子是直角三角形)
          *
         ***
        *****
       *******

以下是两种方法:

print( "方法一" )
row = 1
while row <= 4:
    print( " " * (4 - row) + "*" * (2 * row - 1) + " " * (4 - row) )
 
    row += 1
print( "方法二" )
row = 1
while row <= 4:
    print( ("*" * (2 * row - 1)).rjust( 4 + row - 1, " " ) )
    row += 1
 
print( "方法三升级圣诞树" )
print( "方法三升级圣诞树" )
 
 
def anycount(m):
    row = 1
    while row <= m:
        print( ("*" * (2 * row - 1)).rjust( m + row - 1, " " ) )
        row += 1
    for k in range( 1, row - 1 ):
        print( "*".rjust( (m + row - 1) // 2, " " ) )
 
 
anycount( 10 )

python咋实现输出*为(金字塔)形状,圣诞树形状。_第1张图片

print( "方法四升级圣诞树2" )
 
 
def anycount(m):
    row = 1
    while row <= m:
        print( ("*" * (2 * row - 1)).rjust( m + row - 1, " " ) )
        row += 1
    for k in range( 1, row - 1 ):
        print( "***".rjust( (m + row - 1) // 2 + 1, " " ) )
 
 
anycount( 10 )

python咋实现输出*为(金字塔)形状,圣诞树形状。_第2张图片

你可能感兴趣的:(python咋实现输出*为(金字塔)形状,圣诞树形状。)