PAT (Advanced Level)-1019. General Palindromic Number (20)

PAT (Advanced Level)-1019. General Palindromic Number (20)

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 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


解读

这个问题的考察点很明确:十进制数对任意进制转换的问题。
那么进制的转换的一个思路:(十进制数N,进制基数b)

我们通过辗转相除:对原始数据N每次除以进制基数b,所得余数(N%b)为转换后的最后一位数,再将商(N/b)除以进制基数,余数为倒数第二位,如此循环往复,直至最后商小于进制基数。

从上述算法中我们可以得到一个仅需要对每次的被除数进行修正,即N=N/b,每次将商付给被除数进行操作。
上述算法另外一个需要注意的便是,由于第一次运算得到的是最后一位数,所以输出的时候就需要反序来输出了。

完整代码

#include<iostream>

using namespace std;

void OUTPUT(int transArray[], int FLAG)
{
	for (int i = FLAG - 1; i >= 0; i--)
	{
		if (i != 0)
		{
			cout << transArray[i] << " ";
		}
		else
			cout << transArray[i];
	}
	if (FLAG == 0)
	{
		cout << transArray[0];
	}
}

int main()
{
	int N, b;
	cin >> N >> b;

	int transArray[500];            //数组设置大一些,避免N最大时候,转换2进制时候越界

	int FLAG=0;         //标记转换后的数字有多少位
	int FLAG_XX = 0;            //标记是否为回文数
    
	if (N == 0)         //输入为0时的特殊处理
	{
		transArray[0] = N;
	}
	else            //其他正常输入
	{
		for (; N; FLAG++)
		{
			transArray[FLAG] = N%b;
			N = N / b;
		}

		//判断回文,循环次数可以折半
		for (int i = FLAG - 1; i >= (int)((FLAG - 1) / 2); i--)
		{
			if (transArray[i] != transArray[FLAG - 1 - i])
			{
				FLAG_XX = 1;
			}
		}
	}

    //输出
	if (FLAG_XX == 0)
	{
		cout << "Yes" << endl;
		OUTPUT(transArray, FLAG);
	}
	else if (FLAG_XX == 1)
	{
		cout << "No" << endl;
		OUTPUT(transArray, FLAG);
	}
	return 0;
}

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