Python利用for循环打印星号三角形的案例

for i in range(0,5):
 
 for j in range(i+1):
  if i == 4:
   print("* ",end="")
   continue
  if j == 0 or j == i:
   print("* ",end="")
  else:
   print(" ",end="")
 print()

for i in range(5):
 print("* " * (i+1))

for i in range(5):
 
 print("* "*5)
print()

for i in range(5):
 
 print("* "*5)
print()
for i in range(4):
 if i == 0:
  print("* " * 5)
 if i ==3:
  print("* " * 5)
  continue
 for j in range(5):
  if j == 0:
   print("* ",end=" ")
  if j == 4:
   print("* ")
  else:
   print(" ",end="")

for i in range(5):
 
 print(" "*(4-i),end="")
 print(" * "*(i+1))

#!/usr/bin/python
# -*- coding:utf-8 -*-
#输入合法性检查,必须输入正数,不支持科学计数法
def ispositive(numb):
  try:
    float(numb)
  except:
    return False
  else:
    if float(numb) <= 0:
      return False
    else:
      return True
#直角三角形判断
def ispythagoras(a,b,c):
  if a**2 + b**2 == c**2 or a**2 + c**2 == b**2 or b**2 + c**2 == a**2:
    return True
  else:
    return False

num1 = input("pls enter 1st number:\n")
while not ispositive(num1):
  num1 = input("That's not a valid number. Try again:\n")

num2 = input("pls enter 2nd number:\n")
while not ispositive(num2):
  num2 = input("That's not a valid number. Try again:\n")

num3 = input("pls enter 3rd number:\n")
while not ispositive(num3):
  num3 = input("That's not a valid number. Try again:\n")

num1 = float(num1)
num2 = float(num2)
num3 = float(num3)
#欢迎点评,引用请注明出处
if num1 + num2 > num3 and num2 + num3 > num1 and num1 + num3 > num2:
  if num1 == num2 == num3:
    print("%.2f\n%.2f\n%.2f\n可以组成等边三角形" % (num1,num2,num3))
  elif num1 == num2 or num2 == num3 or num1 == num3:
    if ispythagoras(num1,num2,num3):
      print('%.2f\n%.2f\n%.2f\n可以组成等腰直角三角形' % (num1,num2,num3))
    else:
      print('%.2f\n%.2f\n%.2f\n可以组成等腰三角形' % (num1,num2,num3))
  elif ispythagoras(num1,num2,num3):
    print('%.2f\n%.2f\n%.2f\n可以组成直角三角形' % (num1,num2,num3))
  else:
    print('%.2f\n%.2f\n%.2f\n可以组成普通三角形' % (num1,num2,num3))
else:
  print('%.2f\n%.2f\n%.2f\n不能组成三角形' % (num1,num2,num3))