A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits a_i as the sum of (a_ibⁱ) for i from 0 to k. Here, as usual, 0 <= a_i < b for all i and a_k is non-zero. Then N is palindromic if and only if a_i = a_k-i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

Input Specification:

Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 10⁹ is the decimal number and 2 <= b <= 10⁹ is the base. The numbers are separated by a space.

Output Specification:

For each test case, first print in one line "Yes" if N is a palindromic number in base b, or "No" if not. Then in the next line, print N as the number in base b in the form "a_k a_k-1 ... a₀". Notice that there must be no extra space at the end of output.

Sample Input 1:

27 2

Sample Output 1:

Yes

1 1 0 1 1

Sample Input 2:

121 5

Sample Output 2:

No

4 4 1

思路

进制转换题，检验一个十进制数字用b进制数表示的时候是否是一个回文数。用一个数组保存b进制每一位的数字，然后检查是否对称就行。

代码

#include<iostream>

#include<vector>

using namespace std;

vector<int> digits(30);

int main()

{

    int N,b;

    while(cin >> N >> b)

    {

        int index = 0;

        while( N != 0)

        {

            digits[index++] = N % b;

            N = N/b;

        }

        bool isPal = true;

        for(int i = 0;i < index/2;i++)

        {

            if(digits[i] != digits[index - i - 1])

            {

                isPal = false;

                break;

            }

        }

        if(isPal)

            cout << "Yes" <<endl;

        else

            cout << "No" << endl;

        for(int i = index - 1;i >= 0;i--)

        {

            cout << digits[i];

            if( i != 0)

                cout << " ";

        }

        if(index == 0 )

            cout << 0;

        cout << endl;

    }

}