PAT 1019 General Palindromic Number (20 分)

#include
using namespace std;
int main(){
    int N,b;
    scanf("%d %d",&N,&b);
    int arr[40],index=0;
    while (N>0) {   
        arr[index++]=N%b;
        N=N/b;
    }
    int flag=0;
    for(int i=0;i<index/2;i++){
        if(arr[i]!=arr[index-i-1]){
            printf("No\n");
            flag=1;
            break;
        }
    }
    if(!flag) printf("Yes\n");
    for(int i=index-1;i>=0;i--){
        printf("%d",arr[i]);
        if(i!=0) printf(" ");
    }
    if(index==0) printf("0");
        return 0;
}