用数字1至9组成一个没有重复数字的九位数ABCDEFGHI ,A+AB+BC+CD+DE+EF+FG+GH+HI=HHI，求H和I的值
python代码如何实现所有解。代码如下
A+AB+BC+CD+DE+EF+FG+GH+HI=HHI化简一下，就是A+B+C+D+E+F+G=9H
from itertools import permutations
import pandas as pd
nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]
list = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I']
for i in permutations(nums, len(nums)): #对nums的排列组合结果，类型为tuple
a = pd.Series(i,list)
a1 = a['A'] + a['B'] + a['C'] + a['D'] + a['E'] + a['F'] + a['G']
a2 = 9 * a['H']
if a1 == a2:
print('符合条件的值:A,B,C,D,E,F,G,H,I的顺序', a.values)
