1023 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

Analyze:

    We can use an array to ascertain the property of doubled number. It's easy to double the number with String.

#include<iostream>

using namespace std;

int s[10];
int f[10];

string DoubleNumber(string num){
	string result = "";
	int carry = 0;
	char temp;
	for(int i = num.length() - 1; i >= 0; i--){
		temp = (num[i] - '0') * 2 + '0' + carry;
		carry = 0;
		if(temp > '9'){
			temp -= 10;
			carry = 1;
		}
		result = temp + result;
	}
	if(carry == 1){
		result = '1' + result;
	}
	return result;
}

int main(){
	string num;
	cin >> num;
	for(int i = 0; i < num.length(); i++) s[num[i] - '0']++;
	string result = DoubleNumber(num);
	for(int i = 0; i < result.length(); i++) f[result[i] - '0']++;
	bool flag = true;
	for(int i = 0; i < 10; i++){
		if(f[i] != s[i]){
			flag = false;
			break;
		}
	}
	if(flag){
		cout << "Yes" << endl << result;
	}else{
		cout << "No" << endl << result;
	}
	return 0;
}