PAT 甲级1019. General Palindromic Number (20)

题目：点击打开链接

思路：简单的进制转换，考虑单个数字情况。

代码：

#include<iostream>#include<string>#include<vector>using namespace std;int main(){long long N;int b;cin>>N>>b;vector<int> k;while(N>=b){k.push_back(N%b);N/=b;}k.push_back(N);bool flag=true;for(int i=0;i<=k.size()/2;++i){if(k[i]!=k[k.size()-1-i]){flag=0;break;}}if(flag){cout<<"Yes"<<endl;}elsecout<<"No"<<endl;for(int i=k.size()-1;i>0;--i){cout<<k[i]<<" ";}cout<<k[0]<<endl;system("pause");return 0;}