这题和上题基本一样,但是看翻译时被坑了...
Dual Palindromes
Mario Cruz (Colombia) & Hugo Rickeboer (Argentina)
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.
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 palindromic when expressed in at least two of the bases 2..10. The numbers should be listed in order from smallest to largest.
SAMPLE OUTPUT (file dualpal.out)
26 27 28
=================================
/*
ID: jun41821
PROG: dualpal
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;
ofstream fout ("dualpal.out");
ifstream fin ("dualpal.in");
/* 函数circle用于判断正整数n的d进制数表示形式是否是回文数 */
int circle(int n, int d)
{
int s=0,m=n;
while(m)
{
s=s*d+m%d;
m/=d;
}
return s==n;
}
//函数printd将正整数n由10进制转换为B进制
void printd( int n , int m )
{
char x;int i;
if( n < 0 )
{
fout<<'-';
n = -n;
}
if( n / m )
printd( n / m ,m);
if(n%m<10)
fout<<(n%m);
if(n%m>=10)
{
x='A'+n%m-10;
fout<<x;
}
}
int main()
{
int b,i,j,N,S,k=1,x=0;
fin>>N>>S; //N为判断次数 S为标志
for(i=1;k<=N;i++)
{
S++;
for(b=2;b<=10;b++) //2-10进制判断
if(circle(S,b))
{
x++;
if(x==2)
{
fout<<S<<endl;
x=0;k++;
break;
}
}
x=0;
}
}