12. There are 900 three-digit numbers (100, 101, 999). If these three-digit numbers are printed on cards, with one number per card, some cards, when flipped, still show a three-digit number, such as 198, which when flipped becomes 861 (1 is still considered 1 when flipped); some cards do not, such as 531, which when flipped becomes TEC. Therefore, some cards can be used in two ways, thus the maximum number of cards that can be saved is $\qquad$.
Only the three-digit numbers formed by $0,1,8,6,9$ (the last digit is not 0) will still be three-digit numbers when viewed upside down, and there are $4 \times 5 \times 4=80$ such three-digit numbers. Subtract the 12 numbers that remain the same when viewed upside down: $101,111,181,808$, $818,888,609,619,689,906,916,986$ (note the characteristics of the last 6 numbers). Therefore, the maximum number of cards that can be printed less is $\frac{80-12}{2}=34$.