1019 General Palindromic Number (20 分)

思路

代码一的进制转换用了一个非常繁琐的方法，除法进制转换，所以在第二个代码用了求余进制转换。

回文数，for循环搜索一半判断即可。

代码一

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N,b,n,temp;
    scanf("%d%d",&N,&b);
    int num = 1;

    for(int i =1 ;i<32;i++)
    {
        num = num * b;
        if((N - num)<0)
        {
            n = i-1;
            break;
        }
    }

    int res[32]={0};
    temp = N;
    for (int i = n;i>=0;i--)
    {
        num = 1;
        for(int j = 0;j<i;j++)
        {
            num*=b;
        }
        res[i] =temp  / num;
        temp -= res[i] * num;
    }

    int flag = 1;
    for (int i = n;i>=n/2;i--)
    {
        if(res[i]!= res[n-i])
        {
            flag=0;
            break;
        }
    }
    if(flag)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
        
    for (int i = n;i>=0;i--)
    {
        if(i==0)
            cout<<res[i];
        else
            cout<<res[i]<<' ';
    }

}

代码二

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N,b,n,temp;
    scanf("%d%d",&N,&b);
    int res[32]={0};

    int num = 0;
    do{
        res[num++] = N%b;
        N = N/b;
    }while(N!=0);

    n = num-=1;


    int flag = 1;
    for (int i = n;i>=n/2;i--)
    {
        if(res[i]!= res[n-i])
        {
            flag=0;
            break;
        }
    }
    if(flag)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;

    for (int i = n;i>=0;i--)
    {
        if(i==0)
            cout<<res[i];
        else
            cout<<res[i]<<' ';
    }
}