我用C ++编写了一个显示星号金字塔的程序(见下文),现在我想看看它是如何在Python中完成的,但它并不像我想象的那么容易。
有没有人试过这个,如果是这样的话,你能给我看一些有用的代码吗?
提前致谢。
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答案 0 :(得分:22)
def pyramid(rows=8):
for i in range(rows):
print ' '*(rows-i-1) + '*'*(2*i+1)
pyramid(8)
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pyramid(12)
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答案 1 :(得分:5)
或者你可以试试:
def pyramid(size=8):
for i in range(size):
row = '*'*(2*i+1)
print row.center(2*size)
答案 2 :(得分:3)
你可以像这样使用字符串乘法:
>>> for i in range(size):
... print '%s%s'%(' '*(size-(i-1)),'*'*((i*2)-1))
...
答案 3 :(得分:2)
此代码不是非常pythonic,但它是可读的。看看Hugh Bothewell对紧凑金字塔绘图功能的回答:
def drawPyramid(rows):
result = ''
for i in xrange(rows):
row = ''
row += ' ' * (rows - i - 1)
row += '*' * (2 * i + 1)
result += row + '\n'
return result
print drawPyramid(20)
答案 4 :(得分:2)
我建议使用以下功能:
def pyramid(rows=8):
pyramid_width = rows * 2
for asterisks in range(1, pyramid_width, 2):
print("{0:^{1}}".format("*" * asterisks, pyramid_width))
然后尝试:
pyramid()
或与:
pyramid(4)
答案 5 :(得分:1)
Pyramid, Inverted Pyramid and Diamond Rhombus in Python:
Pyramid
i=1
j=5
while i<=5:
print((j*' ')+i*'* ')
j=j-1
i=i+1
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* *
* * *
* * * *
* * * * *
Inverted Pyramid
i=1
j=5
while i<=5:
print((i*' ')+j*'* ')
j=j-1
i=i+1
* * * * *
* * * *
* * *
* *
*
Diamond Rhombus
i=1
j=5
while i<=5:
print((j*' ')+i*'* ')
while j<=5 & i==5:
print(((j+1)*' ')+(5-j)*'* ')
j=j+1
j=j-1
i=i+1
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* *
* * *
* * * *
* * * * *
* * * *
* * *
* *
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答案 6 :(得分:1)
如果您喜欢列表推导:
> n = 5
> print("\n".join((i*"*").center(n*2) for i in range(1, n*2, 2)))
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答案 7 :(得分:1)
user_input = input("How many rows do you want to print? \n")
def stars(a):
size = int(a)
for i in range(1, size+1):
print(" "*(size-i),"*"*(i*2-1))
stars(user_input)
How many rows do you want to print?
20
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答案 8 :(得分:0)
$ cat tree.py
def line(n, i):
line = ''
for j in range(n - i - 1):
line += ' '
for j in range(2 * i + 1):
line += '*'
for j in range(n - i - 1):
line += ' '
return line
def tree(n):
for i in range(n):
line_ = line(n, i)
print(line_)
def run():
tree(3)
if __name__ == '__main__':
run()
$ python3 tree.py
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$ _
答案 9 :(得分:0)
n=4
m = (2*n)-2
for i in range(0, m):
for j in range(0, m):
print(end=" ")
m = m - 1 # decrementing m after each loop
for j in range(0, i + 1):
# printing full Triangle pyramid using stars
print("* ", end=' ')
print(" ")
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答案 10 :(得分:-1)
你也可以画一个DIAMOND
def pyramid(r):
for i in range(r):
print (" "*(r-i-1) + "*"*(2*i+1))
for i in range(r-1,-1,-1):
print (' '*(r-i-1) + "*"*(2*i+1))
n=int(input("Enter no of rows:"))
pyramid(n)
pyramid(10)
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* * * * * * * * * * *
* * * * * * * * * * * * *
* * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * *
* * * * * * * * * * * * *
* * * * * * * * * * *
* * * * * * * * *
* * * * * * *
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* * *
*
>>>
答案 11 :(得分:-1)
def pyramid(row):
for n in range(row):
print(' '*(n+1)+' '*(2*(row-n))+'x'+'x'*(2*n+1))
pyramid(row=8)
答案 12 :(得分:-1)
虽然我对python非常陌生,所以这就是我解决它的方法:
k=int(input("Enter the number of rows"))
for i in range(1,k):
print(' '*(k-i),'* '*(i))
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* * * * * * *
答案 13 :(得分:-2)
#!/usr/bin/python
for i in range(1,6):
for j in range(1,i+1):
print "*",
print
O/P:
===
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* *
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* * * * *
2)
#!/usr/bin/python
for i in range(1,6):
for j in range(1,7-i):
print "*",
print
O/P:
* * * * *
* * * *
* * *
* *
*
3)
#!/usr/bin/python
for i in range(1,6):
for j in range(1,6-i):
print "",
for k in range(1,i+1):
print "*",
print
O/P:
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* *
* * *
* * * *
* * * * *
4)
#!/usr/bin/python
for i in range(1,6):
for j in range(1,1+i):
print "",
for k in range(1,7-i):
print "*",
print
O/P:
* * * * *
* * * *
* * *
* *
*
5)
#!/usr/bin/python
for i in range(1,6):
for j in range(1,6-i):
print "",
for k in range(1,i+1):
print "*",
print
for i in range(1,5):
for j in range(1,1+i):
print "",
for k in range(1,6-i):
print "*",
print
O/P:
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* *
* * *
* * * *
* * * * *
* * * *
* * *
* *
*