PAT1019. General Palindromic Number

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思路：此题在进制转换的时候记住应该运用整型数组，不要运用字符数组，因为当进制数大于10的时候，可能会出现字母的情况。 

 1 #include<cstdio>
 2 #include<cstring> //回文
 3 int str[1000];    //此题不宜用字符串数组进行存储 因为进制很大的时候可能多出来英文字母
 4 int main(int argc, char *argv[])
 5 {
 6     int n,b;
 7     scanf("%d%d",&n,&b);
 8     int count=0;
 9     if(n==0)
10        str[count++]=0;
11     while(n!=0)
12     {
13         str[count++]=n%b;
14         n/=b;
15     }
16     int flag=true;
17     for(int i=0;i<count/2;i++)
18     {
19         if(str[i]!=str[count-1-i])
20         {
21                flag=false;
22                break;
23          }
24     }
25     if(flag)
26       printf("Yes\n");
27     else
28       printf("No\n");
29     for(int i=count-1;i>=0;i--)
30     {
31         if(i==0)
32            printf("%d\n",str[i]);
33          else
34             printf("%d ",str[i]);
35     }
36
37     return 0;
38 }