1019 General Palindromic Number (20)（20 分）

#include
#include
#include
using namespace std;
int n, b, cnt;
bool ispalindromic(int num[])
{
    for (int i = 0; i < cnt / 2; i++)
    {
        if (num[i] != num[cnt - 1 - i])return false;
    }
    return true;
}
int main()
{
    scanf("%d%d", &n, &b);
    int num[1000];
    do
    {
        num[cnt++] = n%b;
        n /= b;
    } while (n);
    if (ispalindromic(num))printf("Yes\n");
    else printf("No\n");
    reverse(num, num + cnt);
    for (int i = 0; i < cnt; i++)
    {
        printf("%d", num[i]);
        if (i != cnt-1)printf(" ");
    }
    return 0;
}