PAT甲级——A1023 Have Fun with Numbers

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

大数运算， 用字符实现加减

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <string>
 4 using namespace std;
 5 //会溢出，不能使用简单的加减
 6 int main()
 7 {
 8     string a, b = "", res;
 9     cin >> a;
10     int k = 0;
11     for (int i = a.length() - 1; i >= 0; --i)
12     {
13         k = k + (a[i] - '0') + (a[i] - '0');
14         b += k % 10 + '0';
15         k /= 10;
16     }
17     if (k > 0)
18         b += k + '0';
19     res.assign(b.rbegin(), b.rend());    
20     sort(a.begin(), a.end());
21     sort(b.begin(), b.end());
22     if (a == b)
23         cout << "Yes" << endl;
24     else
25         cout << "No" << endl;
26     for (auto c : res)
27         cout << c;
28     cout << endl;
29     return 0;
30 }