1019. General Palindromic Number (20)

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

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最近产量太低，都是有小bug的答案，唯独这道题终于。。。。
其实这道题相当简单，不想说啥了
时间空间都没有限制，如果机试碰到这种题，那估计是八辈子修来的福。。。
直接贴答案吧。。
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#include 
using namespace std;
long resultSize = 0;
const int MAX_BUFF_SIZE = 30;
int buff[MAX_BUFF_SIZE];
void calcBase(long inputVal, long base)
{
	//计算以base为低的inputVal的值
	if(inputVal == 0)
	{
		resultSize = 1;
		buff[0] = 0;
		return;
	}
	resultSize = 0;
	while(inputVal > 0)
	{
		buff[resultSize++] = inputVal % base;
		inputVal /= base;
	}
}
bool isPalindrome()
{
	long limit = resultSize /2;
	for(long i=0; i> a >> b;
	calcBase(a, b);
	if( isPalindrome())
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
	while(--resultSize)
	{
		cout << buff[resultSize] << " ";
	}
	cout << buff[0];
	return 0;
}