1019. General Palindromic number (20)

Time limit MS
Memory Limit 65536 KB
Code length limit 16000 B
Procedures for the award of questions StandardAuthor Chen, Yue

A number that would be the same when it was written forwards or backwards is known as a palindromic number. For example, 1234321 is a palindromic number. All digit numbers is palindromic numbers.

although palindromic numbers is most often considered in the decimal system, the concept of palindromicity can Applie D to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it's written in standard notation with k+1 digits a_{i as the sum of (ai) for I from 0 to K. Here, as usual, 0 <= ai < b for all I and ak is Non-zero. Then N are 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 were supposed to tell if N is a palindromic number in base B.

Input Specification:

Each input file contains the one test case. Each case consists of a non-negative numbers n and b, where 0 <= n <= 9 is the decimal number and^{2 <= b <= 9 is thebase. The numbers is separated by a space.}

Output Specification:

For each test case, the first print in the one line "Yes" if N is a palindromic number in base B, or "No" if not. Then on the next line, print N as the number in base B in the form "a_{k ak-1 ... a0". Notice that there must is no extra space at the end of output.}

Sample Input 1:

27 2

Sample Output 1:

Yes1 1 0 1 1

Sample Input 2:

121 5

Sample Output 2:

No4 4 1

Source: >  

  
 

   
  

  
#include<iostream>
   
  

  
#include<vector>
   
  

  
#include<stdio.h>
   
  

  
#include <algorithm>
   
  

  
#pragma warning(disable:4996)
   
  

  

   
  

  
using namespace std;
   
  

  

   
  

  
vector<int> num;
   
  

  

   
  

  
bool CheckSeq() {
   
  

  
  Span class= "KWD" >for    ( int   i  =    0    I  <   num   size   ()    /   2    +    1    I  ++)    {  
   
  

  
if (num[i] != num[num.size() - i - 1])
   
  

  
return false;
   
  

  
}
   
  

  
return true;
   
  

  
}
   
  

  

   
  

  
int main(void) {
   
  

  
int n, k;
   
  

  
cin >> n >> k;
   
  

  
int yushu;
   
  

  
while (true)
   
  

  
{
   
  

  
yushu = n%k;
   
  

  
num.push_back(yushu);
   
  

  
n = n / k;
   
  

  
if (n == 0)
   
  

  
break;
   
  

  
}
   
  

  

   
  

  
reverse(num.begin(), num.end());
   
  

  

   
  

  
if (CheckSeq() == false) {
   
  

  
cout << "No" << endl;
   
  

  
}
   
  

  
else
   
  

  
cout << "Yes" << endl;
   
  

  

   
  

  
for (int i = 0; i < num.size(); i++) {
   
  

  
cout << num[i];
   
  

  
if (i != num.size() - 1)
   
  

  
cout << " ";
   
  

  
}
   
  

  

   
  

  
return 0;
   
  

  
}
  
 

From for notes (Wiz)

1019. General Palindromic number (20)