1019. General Palindromic Number (20)

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本文介绍了一种判断给定十进制整数N在特定进制b下是否为回文数的方法，并提供了一个C++实现示例。文章详细解释了回文数的概念，即在正读和反读时保持相同的特性，并探讨了这一特性如何应用于不同进制下的自然数。通过具体的输入输出样例展示了算法的工作流程。

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PAT

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

题目大意：输入N和b，对N进行b进制转换，若转换后的数具有对称性，则输出Yes并换行打印每一位的数，反之输出No并换行打印每一位的数，行末无空格。

代码

#include<cstdio>
int main(){
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF){
        int c[1004]={0},flag=0,i=0;
        do{          ///进制转换并逆序存入数组 
            c[i++]=a%b;
            a/=b;
        }while(a!=0);
        int k=i-1;
        for(i=0;i<=k/2;i++)///判断是否是对称数组 
            if(c[i]!=c[k-i])
                flag=1;
        if(flag==0)
            printf("Yes\n");
        else
            printf("No\n");
        for(i=k;i>0;i--)
            printf("%d ",c[i]);
        printf("%d",c[i]);
    }
}