1023. Have Fun with Numbers (20)

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

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

string double_num(string n);
int main()
{
	string n, d_num;
	int key[10] = {0};
	cin >> n;
	for(int i=0; i<n.size(); i++)
	{
		key[n[i]-'0']++;
	}
	d_num = double_num(n);
	if(d_num.size() != n.size())
	{
		cout << "No\n" << d_num << endl;
	}
	else{
		for(int i=0; i<d_num.size(); i++)
		{
			key[d_num[i]-'0']--;
		}
		int flag = 1;
		for(int i=0; i<10; i++)
		{
			if(key[i] != 0)
				flag = 0;
		}
		if(flag)
			cout << "Yes\n" << d_num << endl;
		else
			cout << "No\n" << d_num << endl;
	}
	return 0;
}

string double_num(string n)
{
	int c = 0;
	int temp;
	string d_num;
	for(int i=n.size()-1; i>=0; i--)
	{
		temp = n[i]-'0';
		temp = (temp)*2+c;
		c = temp/10;
		temp %= 10;
		d_num.push_back(temp+'0');
	}
	if(c != 0)
		d_num.push_back(c+'0');
    reverse(d_num.begin(), d_num.end());
	return d_num;
}