1019 General Palindromic Number

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本文探讨了回文数的概念及其在不同进制下的性质，通过实例解析了如何判断一个十进制数在特定进制下是否为回文数，并提供了C++代码实现。

1019 General Palindromic Number

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+1digits ai as ∑i=0k(aibi). 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 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≤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

回文数问题和进制转换属于机考的常规题，要注意根据数值范围来确定解题方法。如进制转换有常规方法（int内）和大数方法（模拟除法）。

#include<iostream>
using namespace std;
int a[101];
int dec2b(int n,int base,int arr[]){
	int idx=0;
	while(n!=0){
		arr[idx]=n%base;
		n/=base;
		idx++;
	}
	return idx;
}
int b2dec4rev(int arr[],int base,int len){
	int ans=0;
	for(int i=0;i<len;i++){
		ans=ans*base+arr[i];
	}
	return ans;
}
int main(void){
	int n,base;
	cin>>n>>base;
	int len=dec2b(n,base,a);
	int reverse = b2dec4rev(a,base,len);
	if(n==reverse) cout<<"Yes"<<endl;
	else cout<<"No"<<endl;

	for(int i=len-1;i>=0;i--){
		if(i==len-1) cout<<a[i];
		else cout<<" "<<a[i]; 
	}

	return 0;
	
}