1019. General Palindromic Number (20)_int numstore=27;-优快云博客

本文介绍了一种方法来判断一个十进制整数是否为特定进制下的回文数，并将其转换为该进制的形式。通过算法实现，能够有效地进行回文数的判断与转换。

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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 <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
using namespace std;

int main()
{
    int n,base;
    cin>>n>>base;
    if(n==0){
        cout<<"Yes"<<endl<<"0";
        return 0;
    }
    int reverseValue  = 0;
    int numStore[30];
    int flag = 0;
    int temp = n;
    while(temp/base!=0 || temp%base!=0){
        numStore[flag++] = temp%base;
        temp = temp/base;
    }
    for(int i=0;i<flag;i++){
        reverseValue += numStore[i]*pow(base,flag-1-i);
    }
    if(reverseValue == n){
        cout<<"Yes"<<endl;
    }else{
        cout<<"No"<<endl;
    }
    for(int i=flag-1;i>0;i--){
        cout<<numStore[i]<<" ";
    }
    cout<<numStore[0];
    return 0;
}