原题：
Dual Palindromes
A number that reads the same from right to left as when read from left to right is called a palindrome. The number 12321 is a palindrome; the number 77778 is not. Of course, palindromes have neither leading nor trailing zeroes, so 0220 is not a palindrome.
The number 21 (base 10) is not palindrome in base 10, but the number 21 (base 10) is, in fact, a palindrome in base 2 (10101).
Write a program that reads two numbers (expressed in base 10):
N (1 <= N <= 15)
S (0 < S < 10000)
and then finds and prints (in base 10) the first N numbers strictly greater than S that are palindromic when written in two or more number bases (2 <= base <= 10).
Solutions to this problem do not require manipulating integers larger than the standard 32 bits.
题意：
一个数字本身可能不是回文数，但是在其他进制下就可能是回文数了。输入两个数N，S，求出大于S的前N个在2-10进制中有两个或以上进制是回文数的数。
进制转换，没有什么技巧，直接上代码：
/* ID:newyear111 PROG: dualpal LANG: C++ */#include <iostream>
#include <fstream>
#include <string>
#include<algorithm>
using namespace std;
const int N=25;char base[N];
char str[N*100];void init(){for(int i=0;i<10;i++){base[i]=char(i+'0');}for(int i=10;i<=20;i++){base[i]=char('A'+i-10);}
// cout<<base<<endl;
}//进制转换函数
int change(int t,int n){int r,num=0;while(t){r=t%n;t/=n;str[num++]=base[r];} str[num]=0;return num;
}//回文数判定函数
bool isOK(int size){for(int i=0;i<size/2;i++){if(str[i]!=str[size-i-1])return false;} return true;
}int main()
{ifstream fin("dualpal.in");ofstream fout("dualpal.out");init();int n,S,size;fin>>n>>S;int i=S+1;while(n){int count=0;for(int j=2;j<=10;j++){size=change(i,j);if(isOK(size)){count++;}}if(count>=2){n--;fout<<i<<endl;}i++;}fin.close();fout.close();return 0;
}