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 <cstdio>
using namespace std;

int main(void)
{
	char num[21];
	scanf("%s", num);
	int digit[10] = {0};
	int i = 0;
	for (; num[i]; ++i)
		++digit[num[i] - '0'];
	int j = 0, remain = 0, temp;
	int num1[22];
	for (i = i -1; i >= 0; --i)
	{
		temp = (num[i] - '0') * 2 + remain;
		--digit[temp % 10];
		num1[j++] = temp % 10 + '0';
		remain = temp / 10;
	}
	if (remain > 0)
	{
		--digit[remain];
		num1[j++] = remain + '0';
	}
	bool isF = true;
	for (int i = 0; i < 10; ++i)
	{
		if (digit[i] != 0)
		{
			isF = false;
			break;
		}
	}
	if (isF)
		printf("Yes\n");
	else
		printf("No\n");
	for (int i = j - 1; i >= 0; --i)
		printf("%c", num1[i]);
	printf("\n");
	return 0;
}