1019 General Palindromic Number

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 (. Here, as usual, 0 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 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 "a​k​​ a​k−1​​ ... a​0​​". 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

题意：

　　给出一个十进制的数字，将这个数字转换成要求的进制，然后判断这个数字是不是回文数字。

思路：

　　本来以为可以用字符串来解决这道题的，但是后面有两个测试点过不了，看了别人的题解之后发现，当进制大于10之后（n % 基数）将是一个两位数字如果这时候再用字符串拼接的话，将会产生两个字符，所以这种方法并不可行。正解应该是，将模后产生的余数保存在一个数组中，最后比较这个数组中的数字是否相等就好了。

Code：

 1 #include <bits/stdc++.h>
 2
 3 using namespace std;
 4
 5 int main() {
 6     int n, b, t;
 7     cin >> n >> b;
 8     vector<int> v;
 9     while (n != 0) {
10         t = n % b;
11         v.push_back(t);
12         n /= b;
13     }
14     if (v.size() == 0) {
15         cout << "Yes" << endl;
16         cout << "0";
17         return 0;
18     }
19     int l = 0, r = v.size() - 1;
20     bool isPalindromic = true;
21     while (l <= r) {
22         if (v[l] != v[r]) {
23             isPalindromic = false;
24             break;
25         }
26         l++;
27         r--;
28     }
29     reverse(v.begin(), v.end());
30     if (isPalindromic) {
31         cout << "Yes" << endl;
32         cout << v[0];
33         for (int i = 1; i < v.size(); ++i) {
34             cout << " " << v[i];
35         }
36     } else {
37         cout << "No" << endl;
38         cout << v[0];
39         for (int i = 1; i < v.size(); ++i) {
40             cout << " " << v[i];
41         }
42     }
43
44     return 0;
45 }