PAT---A1019. General Palindromic Number (20)

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本文介绍了一个程序，用于判断一个十进制非负整数N是否为b进制下的回文数，并将N转换为b进制形式。程序首先读取N和b的值，然后通过除基取余的方法将N转换成b进制数，再检查该数是否为回文数。

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 ai as the sum of (aibi) for i from 0 to k. Here, as usual, 0 <= ai < b for all i and ak is non-zero. Then N is palindromic if and only if ai = ak-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 “ak ak-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

参考代码：`#include
using namespace std;
int main()
{
int a,b;

 int s[100] = {0};
 int c;
 int flag = 1;
 int i=0;
 cin >>a>>b;
 if(a == 0)
 {
      s[i] = 0;
      i = 1;
 }

 else
 {
         while(a)
    {
             s[i] = a%b;
         a = a/b;
             i++;
    }
 }

for(int j=0;j<i/2;j++)
    if(s[j]!=s[i-1-j])
     {
             flag = 0;
         break;
     }

if(flag == 1)
    cout <<"Yes"<<endl;
else
    cout << "No"<<endl;
for(int j=0;j<i;j++)
{
    cout <<s[i-1-j];
    if(j!=i-1)
        cout << " ";
}

 return 0;