PAT(A) 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 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

#include <cstdio>

/*给出两个整数n,b, 问十进制整数n在b进制下是否是回文数
  是则输出Yes, 否则输出No
  再输出n在b进制下的表示
*/

//划重点(2)：判断z[]所存的数是否为回文数。
//即比较位置i与其对称位置num-1-i是否相同，只要有一对位置不同，说明非回文
bool juddge(int z[], int num){
    for(int i=0; i<=num/2; i++)
        if(z[i]!=z[num-1-i])    return false;
    return true;    　　//所有对称位置都相同
}

int main()
{
    int n, b;
    scanf("%d%d", &n, &b);

    //划重点(1)：整数n转换为b进制的数--模板
    int z[40], num=0;       //z[]存放转换结果，num为其位数
    do{
        z[num++]=n%b;       //除基取余（注：y的个位存在z[]的低位下标）
        n=n/b;
    }while(n);              //当n变为0时退出循环

    bool flag=juddge(z, num);   //判断z[]保存的数是否回文
    if(flag==true)  printf("Yes
");
    else    printf("No
");

    for(int i=num-1; i>=0; i--){
        printf("%d", z[i]);
        if(i)   printf(" ");
    }
    return 0;
}