ZOJ 1078

/*判断回文数：
1，进制分解：用模运算和整除运算即可，每一位存在数组c
2，回文判断：将c中按长度len对折，判断对应的位是否相同，有一个不同，就不是回文数*/
#include<iostream>
#include<stdio.h>

using namespace std;

int main()
{
int n;
while (cin >> n&&n)
{
char c[50];
int sign = 0;         //sign=0表示不是回文数
int base[17] = { 0 };
for (int i = 2; i < 17; i++)
{
int m = n;
int len = 0;
//按进制分解
while (m)
{
c[len++] = m%i;
m /= i;
}
sign = 1;
for (int j = 0; j < len/2; j++)
{
if (c[j] != c[len - j - 1])
sign = 0;
}
if (sign)
base[i] = 1;         //在进制i时，是回文
}
int flag = 0;        //没有回文数
for (int i = 2; i < 17; i++)
{
if (base[i] == 1)
flag = 1;
}
if (flag)
{
printf("Number %d is palindrom in basis", n);
for (int i = 2; i < 17; i++)
{
if(base[i]==1)
printf(" %d", i);
}
cout << endl;
}
else
printf("Number %d is not a palindrom\n", n);
}

return 0;
}