推荐|PAT甲级-1019 General Palindromic Number (20分)

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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 $\sum_{i=0}^k (a_i b^i)$ . Here, as usual, 0≤a_i 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 09 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
1

Sample Output 1:

Yes 1 1 0 1 1
1
2

Sample Input 2:

121 5
1

Sample Output 2:

No 4 4 1
1
2

题意：
输入一个十进制数和一个进制数 b
判断这个十进制数转成 b 进制数之后是不是回文数

我的代码：

#include using namespace std; int main() { int N,b; cin>>N>>b; vector<int>changed; while(N>0) { changed.insert(changed.begin(),N%b); N/=b; } bool flag=true; for(int i=0;i<changed.size()/2;i++) if(changed[i]!=changed[changed.size()-i-1])flag=false; if(flag)cout<<"Yes"<<endl; else cout<<"No"<<endl; for(int i=0;i<changed.size();i++) { cout<<changed[i]; if(i!=changed.size()-1)cout<<" "; } return 0; }
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