zoukankan      html  css  js  c++  java
  • PAT甲级——A1019 General Palindromic Number

    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
    
     1 #include <iostream>
     2 #include <vector>
     3 using namespace std;
     4 //此处11进制中的10就是10,而不是a
     5 int main()
     6 {
     7     int N, b;
     8     cin >> N >> b;
     9     vector<int>v1, v2;
    10     while (N)
    11     {
    12         v1.push_back(N%b);
    13         N /= b;
    14     }
    15     v2.assign(v1.rbegin(), v1.rend());
    16     if (v1 == v2)
    17         cout << "Yes" << endl;
    18     else
    19         cout << "No" << endl;
    20     if (v2.size() == 0)
    21         cout << 0;
    22     else
    23     {
    24         cout << v2[0];
    25         for (int i = 1; i < v2.size(); ++i)
    26             cout << " " << v2[i];
    27     }
    28     cout << endl;
    29     return 0;
    30 }
  • 相关阅读:
    苹果将首次采用HTML5直播发布会 狼人:
    Python 3.2 alpha 2发布 狼人:
    下一代Linux文件系统Btrfs走向成熟 狼人:
    Hello! 404 狼人:
    退格回车控制台输入密码
    poj 3233 Matrix Power Series
    地址参考clang: error: linker command failed with exit code 1 (use v to see invocation)
    文本截断JQuery为textarea添加maxlength,并且兼容IE
    代码下载Html5初探视频元素video示例
    c# 限制textbox的输入范围和长度(长度不用maxlength方法)
  • 原文地址:https://www.cnblogs.com/zzw1024/p/11204290.html
Copyright © 2011-2022 走看看