zoukankan      html  css  js  c++  java
  • 1019 General Palindromic Number

    1019 General Palindromic Number (20 分)
     

    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 b2, where it is written in standard notation with k+1 digits ai​​ as (. Here, as usual, 0 for all i and ak​​ is non-zero. Then N is palindromic if and only if ai​​=aki​​ for all i. Zero is written 0 in any base and is also palindromic by definition.

    Given any positive 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 positive numbers N and b, where 0 is the decimal number and 2 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​​ ak1​​ ... 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


    回文数判断,变了进制而已。

     1 #include <bits/stdc++.h>
     2 #define ll long long int
     3 using namespace std;
     4 
     5 ll n, m, k;
     6 ll pos = 0;
     7 ll an[1000];
     8 
     9 int main(){
    10     cin >> n >> m;
    11     while(n){
    12         an[pos++] = n%m ;
    13         n /= m;
    14     }
    15 
    16     bool flag = true;
    17     for(int i = 0 ; i < (pos>>1); i++){
    18         if(an[i] != an[pos-i-1]){
    19             flag = false;
    20             break;
    21         }
    22     }
    23 
    24     if(flag){
    25         cout <<"Yes"<<endl;
    26     }else{
    27         cout <<"No"<<endl;
    28     }
    29     for(int i = pos-1; i >= 0; i--)
    30         printf("%d%c", an[i],i==0?'
    ':' ');
    31     return 0;
    32 }
     
  • 相关阅读:
    centos7 计划任务 定时运行sh
    Nginx负载均衡配置 域名和IP 访问时从java request.getServerName()获取不同问题解决
    windows2012激活方式 2016激活方式 windows10激活方式
    CentOS7 安装Chrome
    使用 go-cqhttp 在龙芯和其他平台搭建qq机器人
    KDE桌面无故特效消失和图标黑底
    SDUST 小学期飞机大战简述
    SDUST 小学期飞机大战简述
    宝塔中极速安装的PHP如何使用AMQP连接RabbitMQ
    Electron登录注册桌面应用源码+安装文件的打包方法
  • 原文地址:https://www.cnblogs.com/zllwxm123/p/11031241.html
Copyright © 2011-2022 走看看