SecretSum 是本次 google 競賽中第二輪淘汰賽的一道分值為 500 分競賽題。事實上，這道題目反而比同輪比賽中的那道 1000 分值的RecurringNumbers 難(RecurringNumbers 的難度水准充其量不過是道初一學生奧數競賽題)。好了，閒話少敘，來看 SecretSum 的題目吧:
一、競賽題目
Problem Statement
We can substitute each digit of a number with a unique letter from ''A'' to ''Z''. This way we can write groups of numbers in a single representation. For example "ABA" can represent any of the following: 101, 151, 343, 767, 929. However, "ABA" cannot mean 555 because each letter must stand for a distinct digit. Furthermore, numbers cannot begin with 0 and thus ''A'' cannot be replaced by 0. Given two such representations num1 and num2 and the result of their summation return the total number of possible combinations of numbers for which the equation holds. If no combinations are possible then return 0.
Definition
Class: SecretSum
Method: countPossible
Parameters: String, String, String
Returns: int
Method signature: int countPossible(String num1, String num2, String result)
(be sure your method is public)
Constraints
- num1, num2, and result will each contain exactly 3 uppercase letters (''A'' - ''Z'').
Examples
0)
"AAA"
"BBB"
"CCC"
Returns: 32
1)
"ABB"
"DEE"
"TTT"
Returns: 112
2)
"ABC"
"ABA"
"ACC"
Returns: 0
Leading zeroes are not allowed.
