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

高精加法：

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int cnt1[10], cnt2[10];

int main() {
	string s, s1 = "";
	cin >> s;
	for (int i = s.size() - 1; i >= 0; i--) {
		cnt1[s[i] - '0']++;
	}
	s = '0' + s;
	s1 = s;
	for (int i = s.size() - 1; i >= 1; i--) {
		int x = s[i] - '0' + s1[i] - '0';
		s1[i] = '0' + x % 10;
		s1[i - 1] += x / 10;
	}
	if (s1[0] != '0') {
		cout << "No" << endl;
		cout << s1;
	} else {
		for (int i = 1; i < s1.size(); i++) {
			cnt2[s1[i] - '0']++;
		}
		bool flag = 1;
		for (int i = 0; i < 10; i++) {
			if (cnt1[i] != cnt2[i]) {
				flag = 0;
				break;
			}
		}
		if (flag) {
			cout << "Yes" << endl;
		} else {
			cout << "No" << endl;
		}
		for (int i = 1; i < s1.size(); i++) {
			cout << s1[i];
		}
	}
	return 0;
}