进制转换+回文字符串+int数组保存进制数 1019 General Palindromic Number (20分)

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本文介绍了一个算法问题，即如何判断一个十进制正整数在给定的进制下是否为回文数。文章详细解释了回文数的概念，并提供了具体的算法实现，包括将十进制数转换为任意进制数的方法，以及如何检查转换后的数字是否为回文。通过实例演示了输入输出格式。

1019 General Palindromic Number (20分)

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 ∑
i=0
k
(a
i
b
i
). 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 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≤10
9
is the decimal number and 2≤b≤10
9
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

解题
将N转化为K进制，查看是否为回文字符串；

注意点
N转化为K进制后，并非一个位置只占一个字符，比如15转为16进制为 15，这里的15占一个字符，故不能用string或char类型保存转换后的数字，必须用int类型保存；

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;

int N,base;

vector<int> Base(int n, int base)
{
	vector<int> res;
	while(n)
	{
		res.push_back(n%base);
		n/=base;
	}
	return res;
}

void isP()
{	
	cin>>N>>base;
	vector<int> res=Base(N,base);
	
	int head=0;
	int tail=res.size()-1;
	bool flag=true;
	while(head<tail)
	{
		if(res[head++]==res[tail--]);
		else {
			flag=false;
			break;
		}
	}
	if(!flag) cout<<"No"<<endl;
	else cout<<"Yes"<<endl;
	
	cout<<res[res.size()-1];
	for(int i=res.size()-2;i>=0;i--)
		cout<<" "<<res[i];
}
 
int main()
{
	isP();
}