1019. General Palindromic Number (20)

题目链接：http://www.patest.cn/contests/pat-a-practise/1019

题目：

1019. General Palindromic Number (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_i as the sum of (a_ibⁱ) for i from 0 to k. Here, as usual, 0 <= a_i < b for all i and a_k is non-zero. Then N is palindromic if and only if a_i = a_k-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 "a_k a_k-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

分析：

就是先把十进制的某个数化为某个进制的数。然后看其是否是回文数，比方27。化为2进制数，是11011，是二进制数。121化为5进制，是441。不是回文

注意点：

看n == 0的情况。

AC代码：

#include<stdio.h>
#include<vector>
using namespace std;
vector<int>V;
int main(void){
 int n, b;
 while (scanf("%d%d", &n, &b) != EOF){
  V.clear();
  if (n == 0){//假设是0的话，任何进制都是回文数字
   puts("Yes");
   printf("0
");
   continue;
  }
  while (n){//用V来存储b进制下的各个位数
   V.push_back(n % b);
   n /= b;
  }
  bool result  = true;;
  for (int i = 0; i < V.size(); i++){
   if (V[i] != V[V.size() - i - 1]){
    result = false;//一有不等的，就不是回文
    break;
   }
  }
  if (result){
   puts("Yes");
  }
  else puts("No");
  for (int i = V.size() - 1; i >= 0; i --){
   if (i == V.size() - 1)printf("%d", V[i]);
   else printf(" %d", V[i]);
  }
  printf("
");
 }
 return 0;
}

截图：

——Apie陈小旭