数字之和与它的取反可以被它们之间的差（绝对值）整除。比如 45 这个数字 (45+54)%|45-54| == 0,要求时输出第满足条件的第 n 个数字的值. 我用下面的代码实现了但是速度很慢,有什么办法可以优化么?

[CupCupFun]

```python
def sum_dif_rev(n):
#反转数字
def revers_num(num):
l = list(str(num))
l.reverse()
return int(''.join(l))
times = 0
i = 45
while True:
if abs(i-revers_num(i)) == 0:
i = i+1
continue
if (i + revers_num(i))%(abs(i-revers_num(i))) == 0:
times = times + 1
#输出第 n 个数字的值
if times == n:
return i
i = i+1
```

[CupCupFun]

突然发现格式有问题,编辑的时候有缩进的,是一定要用 md 编辑才能正确显示缩进么?

[necomancer]

def nextNum():
....i = 0
....while True:
........inv = int(str(i)[::-1])
........if i != inv and ((i+inv) % (i-inv) ==0):
............yield i
........i += 1

for i, num in enumerate(nextNum()):
....if i == n: # stop at nth
........break
....print(num)