题目

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 file 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

我的解决方案

#include <iostream>
#include <algorithm>
#include <string>

//解题思路:
//1.数存入string
//2.用另一个string保存doubling后的结果
//3.sort排序后若一样,则第doubling后的数是原数的一个排列

using namespace std;

string doubling(string num)
{
    int flag = 0; //进位

    for (string::reverse_iterator i = num.rbegin(); i != num.rend(); ++i) {
        int digit = flag + (*i - '0') * 2;
        flag = digit / 10;
        *i = digit % 10 + '0';
    }

    if (flag) {
        num = static_cast<char>(flag + '0') + num;
    }

    return num;
}

int main()
{
    string original, result, tmp;

    cin >> original;
    tmp = result = doubling(original); 
    sort(original.begin(), original.end());
    sort(tmp.begin(), tmp.end());
    cout << (tmp == original ? "Yes" : "No") << endl;
    cout << result << endl;

    return 0;
}