【PAT甲级真题】- General Palindromic Number(20)

题目来源

注意点：

注意格式，末尾不要有空格

Description

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number . For example, 1234321 1234321 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 N>0 N>0 in base b ≥ 2 b≥2 b≥2, where it is written in standard notation with k + 1 k+1 k+1 digits a i a_i ai as ∑ i = 0 k ( a i b i ) \sum_{i=0}^k(a_ib^i) ∑i=0k(aibi). Here, as usual, 0 ≤ a i < b 0≤a_i<b 0≤ai<b for all i i i and a k a_k ak is non-zero. Then N N N is palindromic if and only if a i = a k − i a_i=a_{k−i} ai=ak−i for all i i i. Zero is written 0 0 0 in any base and is also palindromic by definition.

Given any positive decimal integer N N N and a base b b b, you are supposed to tell if N N N is a palindromic number in base b b b.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N N N and b b b , where 0 < N ≤ 10 9 0<N≤10^9 0<N≤109 is the decimal number and 2 ≤ b ≤ 10 9 2≤b≤10^9 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 N N is a palindromic number in base b, or No if not. Then in the next line, print N N N as the number in base b b b in the form " a k a k − 1 ⋯ a 0 a_ka_{k−1}\cdots a_0 akak−1⋯a0". Notice that there must be no extra space at the end of output.

Sample Input 1:

复制代码

27 2

Sample Output 1:

复制代码

Yes
1 1 0 1 1

Sample Input 2:

复制代码

121 5

Sample Output 2:

复制代码

No
4 4 1

题目大意

给定一个整数 N N N ，要求判断它在 b b b 进制下是否是一个回文数，并输出它在 b b b 进制下的数位，以空格分隔。

思路简介

一道简单的模拟题

首先把一个数拆成 b b b 进制，按照如下步骤进行：

数组保存 n % b n\%b n%b
n = n / b n=n/b n=n/b ，回到第一步：

判断回文在上述转换的基础上，用两个指针，一个从最左端开始一个从最右端开始逐位比较即可。

与Reversible Primes很类似，可以参考这篇题解

遇到的问题

无，一遍过

代码

cpp 复制代码

/**
 * https://pintia.cn/problem-sets/994805342720868352/exam/problems/type/7?problemSetProblemId=994805487143337984
 * 模拟
 */
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

void solve(){
    ll n,d;
    cin>>n>>d;
    vector<ll>number;
    while(n){
        number.emplace_back(n%d);
        n/=d;
    }
    int len=number.size();
    int f=0;
    for(int i=0,j=len-1;i<len&&j>=i;++i,--j){
        if(number[i]!=number[j]){
            cout<<"No\n";
            f=1;
            break;
        }
    }
    if(!f)cout<<"Yes\n";
    for(int i=len-1;i>=0;--i){
        cout<<number[i];
        if(i)cout<<' ';
    }
    cout<<'\n';
}

int main(){
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //fstream in("in.txt",ios::in);cin.rdbuf(in.rdbuf());
    int T=1;
    //cin??T;
    while(T--){
        solve();
    }
    return 0;
}