zoukankan      html  css  js  c++  java
  • PAT 甲级 1019 General Palindromic Number(简单题)

    1019. General Palindromic Number (20)

    时间限制
    400 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

    Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits ai as the sum of (aibi) for i from 0 to k. Here, as usual, 0 <= ai < b for all i and ak is non-zero. Then N is palindromic if and only if ai = ak-i for all i. Zero is written 0 in any base and is also palindromic by definition.

    Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

    Input Specification:

    Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 109 is the decimal number and 2 <= b <= 109 is the base. The numbers are separated by a space.

    Output Specification:

    For each test case, first print in one line "Yes" if N is a palindromic number in base b, or "No" if not. Then in the next line, print N as the number in base b in the form "ak ak-1 ... a0". Notice that there must be no extra space at the end of output.

    Sample Input 1:
    27 2
    
    Sample Output 1:
    Yes
    1 1 0 1 1
    
    Sample Input 2:
    121 5
    
    Sample Output 2:
    No
    4 4 1
    
    #include <iostream>
    #include <string.h>
    #include <stdlib.h>
    #include <algorithm>
    #include <stdio.h>
    #include <math.h>
    
    using namespace std;
    int n,b;
    int a[10005];
    int cnt;
    void dfs(int n,int b)
    {
      if(n<b)
      {
        a[cnt++]=n;
        return;
      }
      dfs(n/b,b);
      a[cnt++]=n%b;
    }
    int main()
    {
      while(scanf("%d%d",&n,&b)!=EOF)
      {
        cnt=0;
        dfs(n,b);
        int i=0,j=cnt-1;
        int ans=0;
        while(i<=j)
        {
               if(a[i]!=a[j])
           {
             ans=-1;
             break;
           }
           i++,j--;
        }
        if(ans==-1)
          printf("No
    ");
        else
          printf("Yes
    ");
        for(int i=0;i<cnt;i++)
        {
          if(i==cnt-1)
            printf("%d
    ",a[i]);
          else
              printf("%d ",a[i]);
        }
        
      }
      return 0;
    
    }


  • 相关阅读:
    摄像头距离标定方法研究(得到像素和毫米的转换比)
    mfc通过消息传递参数进行程序间通信
    基于Opencv和Mfc的图像处理增强库GOCVHelper(索引)
    【20160924】GOCVHelper MFC增强算法(4)
    【20160924】GOCVHelper 图像处理部分(3)
    Xamarin Essentials教程使用指南针Compass
    Xamarin Essentials教程使用加速度传感器Accelerometer
    XamarinSQLite教程下载安装SQLite/SQL Server Compact Toolbox
    Xamarin Essentials教程屏幕常亮ScreenLock
    Xamarin Essentials教程振动Vibration
  • 原文地址:https://www.cnblogs.com/dacc123/p/8228609.html
Copyright © 2011-2022 走看看