1019. General Palindromic Number (20)

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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整数在给定的进制下是否为回文数。文章通过示例解释了回文数的概念，并提供了一段C++代码实现，该代码能够将输入的十进制数转换为目标进制形式并检查其是否为回文数。

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 a_i as the sum of (a_ibⁱ) for i from 0 to k. 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 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 "a_k a_k-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

#include<cstdio>
#include<cstring>
using namespace std;
void trans(int x, int n)
{
    int a[1000];
    int cnt = 0;
    while(x)
    {
        a[cnt++] = x%n;
        x /= n;
    }
    int l = 0, r = cnt - 1;
    int flag = 0;
    while(l < r)
    {
        if(a[l] != a[r])
            flag = 1;
        l++;r--;
    }
    if(flag == 1)
    {
        printf("No\n");
    }
    else{
        printf("Yes\n");
    }
    for(int i = cnt - 1; i >= 0; i--)
        {

            printf("%d",a[i] );
            if(i != 0)
                printf(" ");
        }

}
int main()
{
    int N,D;
    scanf("%d%d", &N, &D);
    if(N == 0)
    {
         printf("Yes\n0");
    }
    else{
         trans(N,D);
    }


}