ZOJ Problem Set - 1078 Palindrom Numbers

Palindrom Numbers

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Statement of the Problem

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 Format

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 Format

Your program must print the message 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

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Source: South America 2001

分析：

题意：

有多组测试数据。每组测试数据一个数n（0<n<50000），要求判断n在2~16进制哪些进制下是回文串，并输出全部符合要求的进制数，如果不存在符合条件的进制，输出no。

简单题。对于每个输入，一个for循环判断2~16进制15种情况，每种情况再判断是否是回文串就可以，然后存储回文串的信息。

ac代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char a[20];
bool b[18];
int main()
{
    int n,i,j,k;
    while(scanf("%d",&n)&&n)
    {
        for(i=2;i<=16;i++)
        b[i]=false;
        for(i=2;i<=16;i++)//2~16进制
        {
            int m=n;
            int j=0;
            while(m)
            {
                a[++j]=m%i;
                m/=i;
            }
            for(k=1;k<=j/2;k++)//判断是不是回文串
            {
                if(a[k]!=a[j-k+1])
                break;
            }
            if(k>j/2)
            b[i]=true;
        }
        for(i=2;i<=16;i++)
        if(b[i]) break;
        if(i>16)
        printf("Number %d is not a palindrom\n",n);
        else
        {
            printf("Number %d is palindrom in basis",n);
            for(i=2;i<=16;i++)
            if(b[i])
            printf(" %d",i);
            printf("\n");
        }
    }
    return 0;
}