zoukankan      html  css  js  c++  java
  • A1019 General Palindromic Number一般对称数

    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 Nis 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

    思路:

    将转化进制的数存储在动态数组中,遍历数组的一半判断是否对称,改变标志位,最后输出数组;

     1 #include <iostream>
     2 #include <vector>
     3 using namespace std;
     4 int main() {
     5     int n, b;
     6     vector<int>v;
     7     cin >> n >> b;
     8     while (n != 0) {
     9         v.push_back(n%b);
    10         n = n / b;
    11     }
    12     int flag = 0,len=v.size();
    13     for (int i = 0; i < len / 2 + 1; i++) {
    14         if (v[i] != v[len - 1 - i])
    15             flag = 1;
    16     }
    17     if (flag == 0)cout << "Yes"<<endl;
    18     else cout << "No"<<endl;
    19     for (int i = len - 1; i >= 0; i--) {
    20         if (i != len - 1)cout << " ";
    21         cout << v[i];
    22     }
    23     return 0;
    24 }
    作者:PennyXia
             
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。
  • 相关阅读:
    attr系列
    面对对象中的反射
    Python中的内置函数(比较重要的)
    过滤莫文件夹下所有文件和子文件夹中的文件,并把路径打印出-----面对过程的编程
    python中字典的几个方法介绍
    python中字符串的几个方法介绍
    python中列表与元组
    win7上python2.7连接mysql数据库
    练习-三级菜单
    练习-模拟商城购物车
  • 原文地址:https://www.cnblogs.com/PennyXia/p/12291956.html
Copyright © 2011-2022 走看看