PAT甲级真题 1019 General Palindromic Number (20分) C++实现（注意测试点2、4）

题目

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

思路

将n转为b进制下的数字，存储到vector中，判断数组两端元素是否全部相等即可。可以倒序存储，不影响回文数的判断。

int最大约为2.14 * 10 ^ 9，不必担心int越界。

唯一需要注意的是，b进制下的数字可能大于10，不能用字符串存储，否则测试点2、4无法通过。例如测试用例：
31 20
应输出
No
1 11

代码

#include 
#include 
using namespace std;

int main(){
     
    int n, b;
    cin >> n >> b;
    vector<int> a;
    do{
     
        a.push_back(n % b);  //倒序存储，不影响判断回文数
        n /= b;
    }while (n!=0);
    int len = a.size();
    bool isPalin = true;
    for (int i=0; i<len/2; i++){
     
        if (a[i] != a[len-1-i]){
     
            isPalin = false;
            break;
        }
    }
    if (isPalin){
     
        cout << "Yes" << endl;
    }
    else {
     
        cout << "No" << endl;
    }
    cout << a[len-1];
    for (int i=len-2; i>=0; i--){
     
        cout << " " << a[i];
    }
    cout << endl;
    return 0;
}