1023 Have Fun with Numbers (20)（20 point(s)(cj)

1023 Have Fun with Numbers (20)（20 point(s)）

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>
using namespace std;
int arr[10];
string doub(string s);
int main() {
	string s;
	cin >> s;
	for (int i = 0; i < s.size(); ++i) {
		arr[s[i] - '0']++;
	}
	string rs = doub(s);
	for (int i = 0; i < rs.size(); ++i) {
		arr[rs[i] - '0']--;
	}
	bool f = 1;
	for (int i = 0; i < 10; ++i) {
		if (arr[i] != 0) f = 0;
	}
	if (f)cout << "Yes" << endl;
	else cout << "No" << endl;
	cout << rs << endl;
	system("pause");
	return 0;
}
string doub(string s) {
	string ss(s.rbegin(), s.rend());
	int x,cur=0;
	for (int i = 0; i <ss.size() ; ++i) {
		x = ss[i] - '0';
		x = (x) * 2+cur;
		ss[i] = (x % 10)+'0';
		cur = (x)/10;
	}
	while (cur) {
		ss += (cur % 10) + '0';
		cur /= 10;
	}
	string str(ss.rbegin(), ss.rend());
	return str;
}