<b>谷歌比赛题SecretSum的C++解法</b>[VC/C++编程]

SecretSum 是本次 谷歌 比赛中第二轮淘汰赛的一道分值为 500 分比赛题.事实上,这道标题反而比同轮比赛中的那道 1000 分值的RecurringNumbers 难(RecurringNumbers 的难度水准充其量不过是道初一学生奥数比赛题).好了,闲话少叙,来看 SecretSum 的标题吧:

1、比赛标题

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.

3)

"AAA"

"CDD"

"BAA"

Returns: 32

4)

"TEF"

"FET"

"AAA"

Returns: 12

5)

"ABC"

"ABC"

"BCE"

Returns: 5

We can have the following 5 sums:

124 + 124 = 248

125 + 125 = 250

249 + 249 = 498

374 + 374 = 748

375 + 375 = 750

6)

"AAA"

"AAA"

"BBB"

Returns: 4

We can have the following 4 sums:

111 + 111 = 222

222 + 222 = 444

333 + 333 = 666

444 + 444 = 888

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

标题的意思大致是这样的:数字0-9可以辨别用A-Z中不相同的字母替换,这样便可以用字母构成的字符串代表一个数了.比方"ABA"可以代表: 101, 151, 343, 767, 929.但"ABA"不可以代表555,因为差别的字母必须取差别的数字.此外每个作为开首的字母不取0值.编写一个类 SecretSum,该类中有一个原型为 int countPossible(string num1, string num2, string result)的函数.num1、num2 和 result 就是上面所描写的范例的字符串,此中 result 代表的数为 num1 和 num2 代表的数的和,并且这几个字符串都只含有3个A-Z的字母.返回值为大概获得的组合的总数.
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