1023 Have Fun with Numbers (20 分)

1023 Have Fun with Numbers (20 分)

 

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

#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
#include<string>
using namespace std;
int Array[10];
int main()
{
    string num;
    cin >> num;
    int len = num.size();
    int flag = 0;
    int temp = 0;
    for (int i = len - 1; i >= 0; i--)
    {
        Array[num[i] - '0']++;
        temp = (num[i] - '0') * 2 + flag;
        flag = 0;
        if (temp> 9)
        {
            temp %= 10;
            flag = 1;
        }
        Array[temp]--;
        num[i] = '0' + temp;
    }
    int flag1 = 0;
    for(int i=0;i<10;i++)
        if (Array[i]!=0)
        {
            flag1= 1;
            break;
        }
    if (flag1)
    {
        cout << "No" << endl;
        if (flag)
            cout << "1" << num;
        else
            cout << num;
    }
    else
    {
        cout << "Yes" << endl;
        if (flag)
            cout << "1" << num;
        else
            cout << num;
    }

}