PAT (Advanced Level) 1019. General Palindromic Number (20)

Posted Fighting Heart

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了PAT (Advanced Level) 1019. General Palindromic Number (20)相关的知识,希望对你有一定的参考价值。

简单题。

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;

int s[10000],tot;
int n,b;

bool check()
{
    for(int i=0; i<=tot/2; i++)
    {
        if(s[i]!=s[tot-i-1]) return 0;
    }
    return 1;
}

int main()
{
    scanf("%d%d",&n,&b);
    if(n==0)
    {
        printf("Yes\n");
        printf("0\n");
    }
    else
    {
        tot=0;
        while(n) s[tot++]=n%b,n=n/b;
        if(check()) printf("Yes\n");
        else printf("No\n");
        for(int i=tot-1; i>=0; i--)
        {
            printf("%d",s[i]);
            if(i>0) printf(" ");
            else printf("\n");
        }
    }
    return 0;
}

 

以上是关于PAT (Advanced Level) 1019. General Palindromic Number (20)的主要内容,如果未能解决你的问题,请参考以下文章

PAT (Advanced Level) 1025. PAT Ranking (25)

PTA (Advanced Level)1019 General Palindromic Number

PAT Advanced Level 1044

PAT Advanced Level 1043

PAT Advanced Level 1079

PAT Advanced Level 1095