zoukankan      html  css  js  c++  java
  • A1019. General Palindromic Number (20)

    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
    
     1 // hahaha.cpp : 定义控制台应用程序的入口点。
     2 //
     3 
     4 #include <stdafx.h>
     5 #include <stdio.h>
     6 #include <iostream>
     7 #include <vector>
     8 #include <queue>
     9 #include <map>
    10 #include <string>
    11 #include <cstdio>
    12 #include <set>
    13 #include <algorithm>
    14 #include <string.h>
    15 
    16 using namespace std;
    17 
    18 bool Judge(int z[],int num)
    19     {
    20     for(int i=0;i<=num/2;i++)
    21         {
    22         if(z[i]!=z[num-1-i])
    23             {
    24             return false;
    25             }
    26         }
    27     return true;
    28     }
    29 
    30 
    31 int main() {
    32   
    33     int n,b,z[40],num=0;
    34     scanf("%d %d",&n,&b);
    35     do
    36         {
    37         z[num++]=n%b;
    38         n/=b;
    39         }while(n!=0);
    40 
    41     bool flag=Judge(z,num);
    42     if(flag==true)printf("Yes
    ");
    43     else
    44         printf("No
    ");
    45     for(int i=num-1;i>=0;i--)
    46         {
    47         printf("%d",z[i]);
    48         if(i!=0)printf(" ");
    49         }
    50      return 0;
    51 }
  • 相关阅读:
    零拷贝
    RxJava2源码解析
    一次博客崩溃日志分析
    Spring循环依赖的解决
    解决网络卡顿问题
    软工第一次作业
    3月26-27号训练笔记
    Codeforces Round #708 (Div. 2)题解A,B,C1,C2,E1,E2
    求出所有LIS的可行起点
    2020小米邀请赛决赛补题G,I,J(三DP)
  • 原文地址:https://www.cnblogs.com/ligen/p/4348453.html
Copyright © 2011-2022 走看看