1019 General Palindromic Number (20)（20 分）

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#PAT #1019 #回文数 #甲级

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本文针对PAT甲级1019题——判断十进制整数N在b进制下是否为回文数的问题进行了详细的解析。文章介绍了回文数的概念，并给出了在任意进制中判断一个数是否为回文数的方法。通过示例输入输出，展示了如何使用C++实现该算法。

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题目

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

**题目链接：**https://pintia.cn/problem-sets/994805342720868352/problems/994805487143337984

解题思路

　　**题目大意：**给定一个N和b，求N在b进制下，是否是一个回文数（Palindromic number）。其中，0＜N,b＜＝10^9。
　　使用int作为基本类型就足够了，然后进行进制转换，需要注意的是，b进制比较大，可能得到的不是一个单位数，比如16进制下的10~15。可以用vector<string>进行存储，然后比较。或者用一个的二维数组进行比较也行。

AC代码

/*
** @Brief:No.1019 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-8-03
*/

#include<iostream>
#include<string>
#include<vector>
using namespace std;


int main(){
	int N,radix;
	cin>>N>>radix;
	vector<string> pal;
	int r = N;
	int s;
	do{
		s = r%radix;
		r/=radix;
		pal.push_back(to_string(s));
	}while(r>0);
	
	bool isPal = true; 
	for(int i=0;i<pal.size()/2;i++){
		if(pal[i]!=pal[pal.size()-1-i]){
			isPal = false;
			break;
		}
	}
	cout<<(isPal?"Yes":"No")<<endl;
	cout<<pal[pal.size()-1];
	for(int i=pal.size()-2;i>=0;i--){
		cout<<" "<<pal[i];
	}	
	return 0;
}