PAT A1023 Have Fun with Numbers

题目描述：

　　Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

　　Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

　　Input Specification:
　　Each input contains one test case. Each case contains one positive integer with no more than 20 digits.

　　Output Specification:
　　For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number.

　　Sample Input:
　　1234567899

　　Sample Output:
　　Yes
　　2469135798

参考代码：

 1 /***********************************************
 2 PAT A1023 Have Fun with Numbers
 3 ***********************************************/
 4 #include <iostream>
 5 #include <string>
 6 #include <stack>
 7 #include <map>
 8 
 9 using namespace std;
10 
11 int main() {
12     string num;
13     stack<int> numList;
14     stack<int> doubleList;
15     map<int, int> baseList;
16 
17     cin >> num;
18 
19     for (int i = 0; i < num.size(); ++i) {
20         numList.push(num[i] - '0');
21         baseList[num[i] - '0']++;
22     }
23 
24     int carry = 0, here = 0;
25     while (!numList.empty()) {
26         int n = numList.top();
27         here = (n * 2) % 10 + carry;
28         carry = (n * 2) / 10;
29         doubleList.push(here);
30         numList.pop();
31 
32         baseList[here]--;
33     }
34 
35     if (carry != 0) {
36         doubleList.push(carry);
37         baseList[carry]--;
38     }
39 
40     bool withTheProperty = true;
41     for (map<int, int>::iterator it = baseList.begin(); it != baseList.end(); ++it) {
42         if (it->second != 0) {
43             withTheProperty = false;
44             break;
45         }
46     }
47 
48     cout << (withTheProperty ? "Yes" : "No") << endl;
49 
50     while (!doubleList.empty()) {
51         cout << doubleList.top();
52         doubleList.pop();
53     }
54 
55     return 0;
56 }

View Code

注意事项：

　　1：输入的“正数”已经超过了long long的表示范围，采用string存储再做处理比较恰当。

　　2：输入的“正数” 翻倍之后可能会出现首位仍需进位的情况，因此需要判断一下进位值carry是否为0，如果不为0需要增加空间来存储carry。

　　3：这里采用stack是因为它FILO的特性符合我们从最低位开始翻倍，最后栈内元素出栈的顺序正好就是结果的顺序。当然也可以采用string等其它容器进行处理。

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