Contents

1. 1. Input Specification:
2. 2. Output Specification:
3. 3. Sample Input 1:
4. 4. Sample Output 1:
5. 5. Sample Input 2:
6. 6. Sample Output 2:

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 ai as ∑i=0k(aibi). Here, as usual, 0≤ai<b for all i and ak is non-zero. Then N is palindromic if and only if ai=ak−i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any positive 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 positive 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 “ak ak−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

题目大意：将一个十进制数字转化成b进制数字，并判断其是否是回文数。

分析：将P进制x转换成Q进制存放到z数组中，分为两步：
①将P进制转换成10进制。
②将10进制转换成Q进制。
在本题中给出的已经是十进制了，就直接按第二步走即可。在判断会回文数的时候，只需要判断到length/2-1即可，一旦有不对称的一对，则必定不是回文数。

//进制转换通用代码   
int y = 0,weight = 1，num = 0;//weight是权重
   while (x != 0){//将P进制转换成10进制。
       y = y + ( x % 10 ) * weight;
       x = x / 10;
       weight = weight * P;
   }
   do {//将10进制转换成Q进制
       z[num++] = y % Q;//z[0]~z[nun-1]是Q进制的低位到高位
       y = y / Q;
   }while (y != 0);

#include<iostream>
#include<vector>
using namespace std;
int main(){
    int n,b,length,flag=0;
    vector<int> v;
    cin >> n >> b;
    do {
        v.push_back(n%b);
        n = n / b;
    }while(n!=0);
    length = v.size();
    for (int i=0;i<length/2;i++){
        if (v[i] != v[length-i-1]){
            flag = 1;
            break;
        }
    }
    if (flag == 0) cout << "Yes" << endl;
    else cout << "No" << endl;
    for (int i=length-1;i>=0;i--){
        if (i!=length-1) cout << " ";
        cout << v[i];
    }
    return 0;
}