PAT甲级--1019 General Palindromic Number

题目详情 - 1019 General Palindromic Number (pintia.cn)

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 ∑i=0k​(ai​bi). 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 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<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

错误代码：

起初N的b进制数是用string字符串存储的，没有考虑到b > 10的情况， 测试点2，4不过。

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int N, b;
	cin >> N >> b;
	string s;
	do {
		s += (N % b) + '0';
		N /= b;
	} while (N);
	string ss = s;
	reverse(ss.begin(), ss.end());
	if (ss == s) { cout << "Yes" << endl; }
	else { cout << "No" << endl; }

	for (int i = 0; i < ss.size(); ++i) { 
		if (i != 0) { cout << " "; }
		cout << ss[i]; 
	}

	return 0;
}

正确代码：

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int N, b, flag = 1;
	cin >> N >> b;
	vector<int> v;
	do {
		v.push_back(N % b);
		N /= b;
	} while (N);
	for (int i = 0; i < v.size(); ++i) {
		if (v[i] != v[v.size() - i - 1]) {
			flag = 0;
			break;
		}
	}
	if (flag)cout << "Yes" << endl;
	else  cout << "No" << endl;
	for (int i = v.size() - 1; i >= 0; --i) {
		cout << v[i]; 
		if (i != 0)cout << " ";
	}
	return 0;
}