思路分析：
将n转换为b进制数用数组存储，判断Yes 或No，逆向输出b进制数组
注意：n为0时要做特殊判别
#include"stdio.h"
int main()
{int n, b, num[35], cnt = 0;bool flag = true;scanf("%d %d", &n, &b);if (n==0) {printf("Yes\n");printf("0\n");return 0;}while (n > 0) {num[cnt++] = n % b;n /= b;}for (int i = 0; i < cnt / 2; i++) {if (num[i] != num[cnt - 1 - i]) {flag = false;break;}}if (flag == false) printf("No\n");else printf("Yes\n");for (int i = cnt-1; i >= 0; i--) {printf("%d", num[i]);if (i>0) printf(" ");}printf("\n");return 0;
}题目要求：
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~i~b^i^) 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 <= 10^9^ is the decimal number and 2 <= b <= 10^9^ 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~ … a~0~”. 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