1019 General Palindromic Number (20point(s))_1019 general palindromic number (20 point(s))-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_44970602/article/details/111145277

本文介绍了一个算法，用于判断任意正整数N在给定进制b下是否为回文数。回文数是指顺读和倒读都相同的数。文章详细解释了回文数的概念，并提供了一个具体的算法实现，包括输入输出格式、样例等。

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1019 General Palindromic Number (20point(s))

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 ai as ∑i=0k (ai bi ). 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≤10^9 is the decimal number and 2≤b≤10^9 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

#include<bits/stdc++.h>
using namespace std;
bool isPalindromic(int a[],int n){
    int i=0,j=n-1;
    while(i<=j){
        if(a[i++]!=a[j--])
            return false;
    }
    return true;
}
int main(){
    int n,b;
    cin>>n>>b;
    int a[32];
    int i=0;
    do{
        a[i++]=n%b;
        n/=b;
    }while(n!=0);
    if(isPalindromic(a,i))
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
    for(int j=i-1;j>=0;--j){
        if(j!=i-1)
            cout<<" ";
        cout<<a[j];
    }
    cout<<endl;
}