1019. General Palindromic Number (20)
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题目链接:https://www.patest.cn/contests/pat-a-practise/1019
题目大意:给定正整数N和b,判断N的b进制数是不是回文数
解题思路:
函数Ninb(int n,int b),返回值为n的b进制数的逆置,判断此值就可以判断n的b进制数是否是回文数
函数isPalindromic(vector num),用来判断num是否是回文数
代码如下:
#include <iostream>#include <vector>using namespace std;bool isPalindromic(vector<int> num){//判断数num是否是回文数 int mid=(num.size()-1)/2; for(int i=0;i<=mid;i++){ if(num[i]!=num[num.size()-1-i]) return false; } return true;}vector<int> Ninb(int n,int b){//返回值是n的b进制数的逆置 vector<int> res; if(n==0) res.push_back(n); while(n!=0){ res.push_back(n%b); n=n/b; } return res;}int main(int argc, char const *argv[]){ int n,b; cin>>n>>b; vector<int> res; res=Ninb(n,b); if(isPalindromic(res)) cout<<"Yes"<<endl; else cout<<"No"<<endl; for(int i=res.size()-1;i>=0;i--){ if(i==0) cout<<res[i]<<endl; else cout<<res[i]<<" "; } return 0;}
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