第二层第五题：双基回文数

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Dual Palindromes
Mario Cruz (Colombia) & Hugo Rickeboer (Argentina)

A number that reads the same from right to left as when read from left toright is called a palindrome. The number 12321 is a palindrome; the number77778 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 Nnumbers strictly greater than S that are palindromic when written in two ormore number bases (2 <= base <= 10).

Solutions to this problem do not require manipulating integers larger thanthe standard 32 bits.

PROGRAM NAME: dualpal

INPUT FORMAT

A single line with space separated integers N and S.

SAMPLE INPUT (file dualpal.in)

3 25

OUTPUT FORMAT

N lines, each with a base 10 number that is palindromicwhen expressed in at least two of the bases 2..10. The numbers should be listedin order from smallest to largest.

SAMPLE OUTPUT (file dualpal.out)

题目大意：如果一个数字在两个或更多进制下（2≤进制数≤10）均为回文数，则称该数为双基回文数。题目给出两个数字：S（0<S<10000）和N(1≤N≤15)，求出大于S的连续N个双基回文数。

枚举数字，然后枚举进制，然后转换进制，然后判断，然后输出，然后。。。AC

代码如下：

/*
ID:su1q2d2
LANG:C++
TASK:dualpal
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#define maxl 50
#define lint long long
using namespace std;
void cn(char ans[],int x,int b)
{
	int i=0;
	while(x>0){
		ans[i]=x%b+'0';
		x/=b;
		i++;
	}
	ans[i]='\0';
	return ;
}
bool pal(char x[])
{
	int n;
	int i;
	n=strlen(x);
	for(i=0;i<n;i++){
		if(x[n-i-1]!=x[i]) return false;
	}
	return true;
}
int main()
{
	freopen("dualpal.in","r",stdin);
	freopen("dualpal.out","w",stdout);
	int i;
	int j;
	int c;
	int s;
	int n;
	int is;
	char afc[maxl];
	scanf("%d",&n);
	scanf("%d",&s);
	c=0;
	for(i=s+1;c<n;i++){
		is=0;
		for(j=2;j<=10;j++){
			cn(afc,i,j);
			if(pal(afc)) is++;
			if(is>=2) break;
		}
		if(is>=2){
			printf("%d\n",i);
			c++;
		}
	}
	return 0;
}