fjnu 1827 Palindrom Numbers

本文介绍了一种算法,该算法能够检测一个给定的整数是否为2到16进制之间的任何一种进制下的回文数。通过将数字从十进制转换为指定进制,并检查其是否为回文数来实现这一目标。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

 

Description

We say that a number is a palindrom if it is the sane when read from left to right or from right to left. For example, the number 75457 is a palindrom.

Of course, the property depends on the basis in which is number is represented. The number 17 is not a palindrom in base 10, but its representation in base 2 (10001) is a palindrom.

The objective of this problem is to verify if a set of given numbers are palindroms in any basis from 2 to 16.

Input

Several integer numbers comprise the input. Each number 0 < n < 50000 is given in decimal basis in a separate line. The input ends with a zero.

Output

Your program must print the mes

Source:#include<iostream>
using namespace std;

int n;

int Palindrom(int x)
{
    
int temp[100];
    
int i=0,j=0;
    
int t;
    t
=n;
    
while(t!=0)
    
{
        temp[j
++]=t%x;
        t
=t/x;
    }

    j
--;
    
while(i<j)
    
{
        
if(temp[i++]!=temp[j--]) return 0;
    }

    
return 1;
}


int main()
{
//    freopen("fjnu_1827.in","r",stdin);
    int m,a[20];
    
int i;
    cin
>>n;
    
while(n!=0)
    
{
        m
=0;
        
for(i=2;i<=16;i++)
        
{
            
if(Palindrom(i)) a[++m]=i;
        }

        
if(m>0)
        
{
            cout
<<"Number "<<n<<" is palindrom in basis ";
            
int j;
            
for(j=1;j<=m-1;j++)
                cout
<<a[j]<<" ";
            cout
<<a[j]<<endl;
        }

        
else
            cout
<<"Number "<<n<<" is not a palindrom"<<endl;
        cin
>>n;
    }

    
return 0;
}


sage Number i is palindrom in basis where I is the given number, followed by the basis where the representation of the number is a palindrom. If the number is not a palindrom in any basis between 2 and 16, your program must print the message Number i is not palindrom.

Sample Input

17
19
0

 

Sample Output

Number 17 is palindrom in basis 2 4 16
Number 19 is not a palindrom
KEY:先进制转化,然后判断;
 

 

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值