1. (3 points) In the addition equation on the right, each Chinese character represents a non-zero digit, and different characters represent different digits. When the equation holds, 贺 + 新 + 春 = ( )
放 鞭 炮
A. 24
B. 22
C. 20
D. 18
【Analysis】According to the problem "放鞭炮" + "迎龙年" = "贺新春", and because the sum of the numbers 1 to 9 is 45, we can solve the problem by discussing the four aspects of both the tens and units place adding up to a carry, only the units place adding up to a carry, only the tens place adding up to a carry, and no carry at all, based on the rules of addition.

【Solution】Solution: (1) Assuming both the tens and units place add up to a carry, we get: 炮 + 年 = 10 + 春, 鞭 + 龙 = 10 + 新 - 1 = 9 + 新, 放 + 迎 = 贺 - 1, then 炮 + 年 + 鞭 + 龙 + 放 + 迎 = 10 + 春 + 9 + 新 + 贺 - 1 = 贺 + 新 + 春 + 18, so 放 = 鞭 + 炮 + 迎 + 龙 + 年 + 贺 + 新 + 春 = 2 (贺 + 新 + 春) + 18 = 45, which means 贺 + 新 + 春 = 27 / 2, which does not meet the problem's requirements;
(2) Assuming only the units place adds up to a carry, then 炮 + 年 = 10 + 春, 鞭 + 龙 = 新 - 1, 放 + 迎 = 贺, then 炮 + 年 + 鞭 + 龙 + 放 + 迎 = 10 + 春 + 新 - 1 + 贺 = 贺 + 新 + 春 + 9, so 放 = 鞭 + 炮 + 迎 + 龙 + 年 + 贺 + 新 + 春 = 2 (贺 + 新 + 春) + 9 = 45, which means 贺 + 新 + 春 = 18, which meets the problem's requirements;
(3) Assuming only the tens place adds up to a carry, then 炮 + 年 = 春, 鞭 + 龙 = 10 + 新, 放 + 迎 = 贺 - 1, then 炮 + 年 + 鞭 + 龙 + 放 + 迎 = 春 + 10 + 新 + 贺 - 1 = 贺 + 新 + 春 + 9, so 放 = 鞭 + 炮 + 迎 + 龙 + 年 + 贺 + 新 + 春 = 2 (贺 + 新 + 春) + 9 = 45, which means 贺 + 新 + 春 = 18, which meets the problem's requirements;
(4) Assuming no carry at all, then 炮 + 年 = 春, 鞭 + 龙 = 新, 放 + 迎 = 贺, then 炮 + 年 + 鞭 + 龙 + 放 + 迎 = does not meet the problem's requirements.

In summary, 贺 + 新 + 春 = 18.
Therefore, the answer is: D.