1019 General Palindromic Number (20分)

 1 #include<iostream>
 2 #include<vector>
 3 #include<cstdio>
 4 using namespace std;
 5 int main(){ 
 6     int n,b;
 7     cin>>n>>b;
 8     int digit[10000]={0};
 9     int len=0;
10     while(n!=0){//十进制转换为base进制 
11         digit[len++]=n%b;
12         n=n/b;
13     }
14     int lo=0,hi=len;
15     int flag=1;
16     while(lo<hi){
17         if(digit[lo]!=digit[hi-1])
18           {flag=0;
19               break;
20           }
21           lo++;
22           hi--;
23     } 
24     if(flag)
25       cout<<"Yes"<<endl;
26     else
27       cout<<"No"<<endl;
28     for(int i=len-1;i>=0;i--){    
29         if(i==0)
30           cout<<digit[i];
31         else
32           cout<<digit[i]<<" ";
33     }   
34     return 0;
35 }