1019. General Palindromic Number (20)

400 ms

65536 kB

16000 B

Standard

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

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;
}