zoukankan      html  css  js  c++  java
  • [PAT] 1019 General Palindromic Number (20 分)Java

    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>0N > 0N>0 in base b≥2b ge 2b2, where it is written in standard notation with k+1k+1k+1 digits aia_iai​​ as ∑i=0k(aibi)sum_{i=0}^k (a_ib^i)i=0k​​(ai​​bi​​). Here, as usual, 0≤ai<b0 le a_i < b0ai​​<b for all iii and aka_kak​​ is non-zero. Then NNN is palindromic if and only if ai=ak−ia_i = a_{k-i}ai​​=aki​​ for all iii. Zero is written 0 in any base and is also palindromic by definition.

    Given any positive decimal integer NNN and a base bbb, you are supposed to tell if NNN is a palindromic number in base bbb.

    Input Specification:

    Each input file contains one test case. Each case consists of two positive numbers NNN and bbb, where 0<N≤1090 < N le 10^90<N109​​ is the decimal number and 2≤b≤1092 le b le 10^92b109​​ is the base. The numbers are separated by a space.

    Output Specification:

    For each test case, first print in one line Yes if NNN is a palindromic number in base bbb, or No if not. Then in the next line, print NNN as the number in base bbb in the form "aka_kak​​ak−1a_{k-1}ak1​​ ... a0a_0a0​​". 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 package pattest;
     2 
     3 import java.io.BufferedReader;
     4 import java.io.IOException;
     5 import java.io.InputStreamReader;
     6 
     7 /**
     8  * @Auther: Xingzheng Wang
     9  * @Date: 2019/2/20 22:41
    10  * @Description: pattest
    11  * @Version: 1.0
    12  */
    13 public class PAT1015 {
    14 
    15     public static void main(String[] args) throws IOException {
    16         BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    17         while (true) {
    18             String[] in = reader.readLine().split(" ");
    19             int num = Integer.parseInt(in[0]);
    20             if (num < 0) break;
    21             int indax = Integer.parseInt(in[1]);
    22             if (isprime(num) && isprime(reve(num,indax))) {
    23                 System.out.println("Yes");
    24             }else{
    25                 System.out.println("No");
    26             }
    27         }
    28     }
    29 
    30     private static int reve(int num, int indax) {
    31         StringBuilder sb = new StringBuilder(Integer.toString(num, indax)).reverse();
    32         return Integer.parseInt(sb.toString(), indax);
    33     }
    34 
    35     private static boolean isprime(int num){
    36         if(num == 1) return false;
    37         for (int i = 2; i <= Math.sqrt(num); i++) {
    38             if (num % i == 0)
    39                 return false;
    40         }
    41         return true;
    42     }
    43 }
  • 相关阅读:
    AOSP 设置编译输出目录
    android stadio 编译报错:download fastutil-7.2.0.jar
    Ubuntu adb 报错:no permissions (user in plugdev group; are your udev rules wrong?);
    Ubuntu 18启动失败 Started Hold until boot procss finishes up
    算法---------两数之和
    Windows 显示环境变量
    Android ObjectOutputStream Serializable引发的血案
    (AOSP)repo checkout指定版本
    如果看懂git -help
    Android stado 运行项目,apk does not exist on disk.
  • 原文地址:https://www.cnblogs.com/PureJava/p/10498011.html
Copyright © 2011-2022 走看看