1019 General Palindromic Number （20 分）(回文数)

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本文介绍了一种算法，用于判断一个十进制正整数在特定的进制下是否为回文数。通过将数字转换为指定进制并检查其是否与其反转相同来实现。文章提供了一个C++代码示例，演示了如何实现这一功能。

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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

as ∑

i=0

). Here, as usual, 0≤a

<b for all i and a

is non-zero. Then N is palindromic if and only if a

k−i

for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any positive 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 positive numbers N and b, where 0<N≤10

is the decimal number and 2≤b≤10

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−1

... a

". 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:

4 4 1

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
using namespace std;
int main(){
    int n,b,arr[1000],len=0;
    cin>>n>>b;
    int temp=n;
    while(n!=0){
        arr[len++]=n%b;
        n/=b;
    }
    for(int i=0;i<len;i++)
        n=arr[i]+n*b;       //倒序相加，如果是回文数，则与原来相等
    if(n==temp) cout<<"Yes\n";
    else cout<<"No\n";
    cout<<arr[len-1];
    for(int i=len-2;i>=0;i--)
        cout<<" "<<arr[i];
    return 0;
}