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:

Yes2469135798

#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;elsecout << "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;}