1019. General Palindromic Number (20)

题目连接：https://www.patest.cn/contests/pat-a-practise/101

原题如下：

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
这道题应该属于简单题了……一开始我还担心数会很大，得用数组，后来发现int足矣……
就是将十进制数转换为任意进制，然后判断是否为回文数。注意下0的情况即可。

#include<stdio.h>
#define MAXN 1005
int main()
{
     unsigned int N,b,n;
     scanf("%u%u",&N,&b);
     unsigned int Traverse[MAXN]={0};
    int i=0;
    n=N;
    if (n!=0)
    {
    while (N)
    {
        Traverse[i++]=N%b;
        N/=b;
    }
    int flag=1;
    int k;
    k=--i;
    for (i=0;i<=k/2;i++)
    {
        if (Traverse[i]!=Traverse[k-i])flag=0;
    }

    if (flag)printf("Yes
");
    else printf("No
");
    for (i=k;i>=0;i--)
    {
        printf("%u",Traverse[i]);
        if (i!=0)printf(" ");
    }
    }
    else if (n==0)
    {
        printf("Yes
");
        printf("0");
    }
    return 0;
}