就是用0到9这几个数，组成两个三位数和一个四位数，使得两个三位数相加等于那个四位数
这十个数字都要用到并且不能重复
我就简单粗暴的遍历
然后一点点排除
过程那可太，难了
看能不能优化一下下
求佬们指点指点
for a in range(100,1000):
for b in range(100,1000):
c = a + b
if 999 <= c <= 2000:
a_gewei = a % 10
a_shiwei = a // 10 % 10
a_baiwei = a // 100
b_gewei = b % 10
b_shiwei = b // 10 % 10
b_baiwei = b // 100
c_gewei = c % 10
c_shiwei = c // 10 % 10
c_baiwei = c // 100 % 10
c_qianwei = c // 1000
if a_gewei != a_shiwei and c_baiwei != b_baiwei and c_qianwei != b_baiwei and c_baiwei != c_gewei and c_gewei != a_gewei and c_baiwei != a_gewei and c_qianwei != a_shiwei and c_qianwei != a_baiwei and c_qianwei != a_gewei and c_qianwei != c_shiwei and c_baiwei != a_baiwei and c_gewei != a_shiwei and c_baiwei != a_shiwei and c_baiwei != b_shiwei and c_gewei != c_qianwei and c_gewei != b_shiwei and c_shiwei != a_gewei and a_shiwei != a_baiwei and a_gewei != a_baiwei and a_gewei != b_gewei and a_gewei != b_shiwei and a_gewei != b_baiwei and a_shiwei != b_baiwei and a_baiwei != b_shiwei and a_shiwei != b_shiwei and a_shiwei != c_shiwei and c_gewei != b_baiwei and c_gewei != a_baiwei and c_shiwei != b_baiwei and c_shiwei != a_baiwei and c_shiwei != a_shiwei and c_shiwei != b_shiwei and c_qianwei != b_shiwei and a_baiwei != b_baiwei and a_baiwei != b_gewei and a_shiwei != b_gewei and b_gewei != b_shiwei and b_shiwei != b_baiwei and b_baiwei != b_gewei and b_gewei != c_gewei and b_gewei != c_shiwei and b_gewei != c_baiwei and b_gewei != c_qianwei and c_gewei != c_shiwei and c_shiwei != c_baiwei and c_baiwei != c_qianwei:
print(f"{c} = {a} + {b}")
https://blog.csdn.net/caolaosanahnu/article/details/14005137