我正在try 这个问题，我只需要使用下面给出的代码来制作这个三角形图案:
*
***
*****
*******
*********
以下是允许使用的代码:
count += 1
star = "*"
continue
count = 10
elif:
print (star,end="")
print()
print (star)
break
i % 2 == 0
i % 2 == 1
这是我设法做的，我似乎找不到一种方法来获得上面的输出与代码:
count=10
star="*"
for i in range (count):
if i%2==0:
print(star)
else:
for x in range (i):
print(star,end="*")
count+=1
好的，首先要做的是:你在第一行中的定义是正确的:
count=10
star="*"
for i in range(count):
如果你仔细观察这个图案，你就会注意到，如果你仔细观察这个图案，每一条线都是由奇数颗星星组成的:1，3，5……所以从根本上说，你想跳过每一次有偶数颗星的迭代，这就是continue的意义所在，把你带到下一个迭代循环.
但是请记住，Pythonrange迭代从0开始，所以每一行都比它的索引多包含一个星号.总而言之，you want to print even lines (with an odd number of stars) and skip odd lines (with an even number of stars)项:
if (i%2 == 1): # if you're looking at an odd line (with an even number of stars)
continue # skip this iteration
else: # we're now sure we're on an even line
现在我们只有偶数线了.对于这些行中的每一行，我们希望保留与行索引(i索引)+1一样多的星号.当使用print方法时，end的缺省值是\n，这是换行符，所以当您使用print时，默认情况下它打印一行并转到下一行.print(star, end="")指令将通过强制代码保持在同一行而不是转到新行来帮助我们.还记得我们说过我们想打印(i+1)颗星星吗？
for x in range(i):
print(star,end="") #print our star and prevents the new line
print(star) #once we printed our i stars, we print the i+1 and go to a new line
总而言之:
count=10
star="*"
for i in range (count):
if (i%2 == 1): # if you're looking at an odd line (with an even number of stars)
continue # skip this iteration
else: # we're now sure we're on an even line
for x in range(i):
print(star,end="") #print our star and stay on the same line
print(star) #once we printed our i stars, we print the i+1 and go to a new line