011
0.False1.3
2.6
3.0
4.True
5.1<2 and 2>3 and 3<4 and 4<5
-------------------------------------dds
0.
steps = 7
i = 1
FIND = False
while i < 1000:
if i%2 ==1 and i%3 == 2 and i%5 == 4 and i%6 == 5 and i%7 == 0:
FIND = True
steps = i
break
else:
FIND = False
i = i + 1
if FIND == True:
print('阶梯数是：', steps)
else:
print('在程序限定的范围内找不到答案！')
**************************************
改错：
0.报错
3.false
4.false，注意，这个链式比较，拆出来是 1 == 2 and 2 < 3，而不是先 (1 == 2) < 3，得到的是两个截然不同的结果。
------------------
dds
0.
steps = 7
i = 1
FIND = False
while i < 100:
if (steps % 2 == 1) and (steps % 3 == 2) and (steps % 5 == 4) and (steps % 6 == 5):
FIND = True
break
else:
steps = 7 * (i + 1)
print(i,steps)
i = i + 1
if FIND == True:
print('阶梯数是：', steps)
else:
print('在程序限定的范围内找不到答案！')
首先steps是7 的倍数，其次，每次增加7，看是否匹配。
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