import itertools
l=[]
for i in range(2,10):
l.append(i)
x=list(itertools.permutations(l,6))
x1=[]
x2=[]
for i in range(len(x)):
x1=x[i]
if(int(x1[0])+int(x1[1])==int(x1[2]*10)+int(x1[3])-int(x1[4])==10+int(x1[5])):
x2.append(x1)
for i in range(len(x2)):
for j in range(i):
if(x2[i][2:5]==x2[j][2:5]):
print("{}+{}={}{}-{}=1{}".format(x2[i][0],x2[i][1],x2[i][2],x2[i][3],x2[i][4],x2[i][5]))
运用标准库itertools提供的排列函数permutations()函数，将2-9这八个数字取出需要的六位进行一个全排列组合
x=list(itertools.permutations(l,6))
进行一个与算式匹配的预算，将这些排列存于列表x2中，便于下面的去重
for i in range(len(x)): x1=x[i] if(int(x1[0])+int(x1[1])==int(x1[2]*10)+int(x1[3])-int(x1[4])==10+int(x1[5])): x2.append(x1)
因为8+9后续还会出现9+8 所以要去重，因为无论是8+9还是9+8，后面的答案都是23-6，重复的算式中间三位一定是相等的，切片比较这三位，然后去重输出
for i in range(len(x2)):
for j in range(i):
if(x2[i][2:5]==x2[j][2:5]):
print("{}+{}={}{}-{}=1{}".format(x2[i][0],x2[i][1],x2[i][2],x2[i][3],x2[i][4],x2[i][5]))
结果
8+6=23-9=14 8+7=24-9=15 9+6=23-8=15 9+7=24-8=16 9+8=23-6=17