PAT Advanced 1019 General Palindromic Number (20分)

Posted 2021-03-13 littlepage

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 (. Here, as usual, 0 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 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 is the decimal number and 2 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?? ... a?0??". 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

 

鸣谢网友“CCPC拿不到牌不改名”修正数据！

题意：判断一个树在n进制下，是否是回文数。

我们使用两个双端队列进行存储回文。

一个从队头插入，一个从队尾插入，然后遍历一遍，比较一下两个队列的数据是否一样。

如果一样输出Yes，否则输出No。最后输出一下这个数。

#include <iostream>
#include <deque>
using namespace std;
int main(){
    int N, R;
    cin >> N >> R;
    deque<int> d, dr;
    while(N != 0){
        d.push_back(N % R);
        dr.push_front(N % R);
        N /= R;
    }
    bool right = true;
    for(int i = 0; i < d.size(); i++)
        if(d[i] != dr[i]) right = false;
    cout << (right ? "Yes" : "No") << endl;
    for(int i = 0; i < dr.size(); i++)
        if(i != 0) cout << " " << dr[i];
        else cout << dr[i];
    return 0;
}