1023. Have Fun with Numbers (20)

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本文介绍了一种检测方法，用于判断一个给定的数字在翻倍后是否仅由其原始数字的不同排列组成。通过逐位翻倍并检查翻倍后的数字是否包含原始数字的相同数量的每一位数字来实现这一目标。

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;
}