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

 #include <iostream>
 #include <algorithm>
 #include <string>
 using namespace std;
 //会溢出，不能使用简单的加减
 int main()
 {
     string a, b = "", res;
     cin >> a;
     int k = ;
     for (int i = a.length() - ; i >= ; --i)
     {
         k = k + (a[i] - '') + (a[i] - '');
         b += k %  + '';
         k /= ;
     }
     if (k > )
         b += k + '';
     res.assign(b.rbegin(), b.rend());
     sort(a.begin(), a.end());
     sort(b.begin(), b.end());
     if (a == b)
         cout << "Yes" << endl;
     else
         cout << "No" << endl;
     for (auto c : res)
         cout << c;
     cout << endl;
     return ;
 }