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 = 0;
    for (int i = a.length() - 1; i >= 0; --i)
    {
        k = k + (a[i] - '0') + (a[i] - '0');
        b += k % 10 + '0';
        k /= 10;
    }
    if (k > 0)
        b += k + '0';
    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 0;
}