1019. General Palindromic Number (20)

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

算法分析：此题就是把回文串扩展到任意进制的自然数。主要方法就是先将输入转换为相对应进制的数，把每一位保存在整型数组中，接着判断是否是回文，最后反向输出数字。

#include <stdio.h>#include <stdlib.h>int main(){    int n, b, i = 0, j, flag = 1, tmp;    int num[100] = {0};    scanf("%d %d", &n, &b);    if(n == 0) //注意为0的情况    {        printf("Yes\n0");        return 0;    }    while(n) //进制转换    {        num[i++] = n % b;        n /= b;    }    tmp = i;    for(j = 0, i--; j < i; j++, i--) //判断回文    {        if(num[j] != num[i])        {            flag = 0;            break;        }    }    if(flag) //输出判断结果        printf("Yes\n");    else        printf("No\n");    for(i = tmp - 1; i >= 0; i--) //输出数字    {        printf("%d", num[i]);        if(i != 0)            printf(" ");    }    return 0;}