wrw5192 发表于 2020-5-14 10:41
人家说了不许用print了,怎么这么多还用。
就是提醒一下
是说 不许直接 print() 输出{:10_245:}
for i in range(10,100):
if str(i)=="3"or str(i)=="3" or i%3==0:
continue
else:
print(i,end=" ")
1
for i in range(10,100):
if i%3 == 0 or ('3' in str(i)):
continue
else:
print(i,end=',')
for i in range(10, 99):
if not str(i).__contains__(str(3)) and i % 3 != 0:
print(i, end=" ")
def no3():
lst=[]
for i in range(10,100):
if (i//10)%3!=0 and (i%10)%3!=0 and i%3!=0:
lst.append(i)
print(lst)
no3()
看了你的运行效果才发现你这题目会产生歧义,我理解成:十位 & 个位上(不为 3且 不是 3 的倍数){:10_266:}
liuzhengyuan 发表于 2020-5-11 19:45
我这里报错?
那你改成 print()
难
""" 十位&个位上不为3且不是3的倍数的两位数"""
for N in range(10,100):
if N %3 != 0 and not("3" in str(N)):
print(N)
for i in range(10,100):
if i%3 != 0 and i//10 !=3:
print(i, end=' ')
else:
pass
{:10_277:}
import re
a = r'3\d{1}'
b = r'\d{1}3'
for i in range(10, 100):
if re.findall(a, str(i)) or re.findall(b, str(i)) or i % 3 == 0:
continue
else:
print(i, end=' ')
sunrise085 发表于 2020-5-11 14:44
赞
for i in range(10,100):
if i%3!=0 and int(str(i))!=3 and int(str(i))!=3:
print(i,end=",")
WangJS 发表于 2020-5-11 15:39
zhan
怎么这么久都还没更新呀
1
zq015 发表于 2020-5-11 21:02
for i in range(1,10):
for j in range(0,10):
ifi !=3 and j != 3 and (i*10+j)%3 != 0:
...
for i in range(10,100):
if i%3!=0 and i%10!=3 and i//10!=3:
print(i,end=' ')
留