想不到更有效的方法，數(shù)據(jù)量不大，直接暴力
#include <iostream>
#include <fstream>
using namespace std;
ifstream fin("dualpal.in");
ofstream fout("dualpal.out");
#ifdef _DEBUG
#define out cout
#define in cin
#else
#define out fout
#define in fin
#endif
int buf[33];
bool isPal(int num, int base)
{
int cnt = 0;
while(num!=0){
buf[cnt++] = num%base;
num/=base;
}
int max = (cnt-1)/2;
for(int i=0;i<=max;++i){
if(buf[i]!=buf[cnt-1-i])
return false;
}
return true;
}
bool duaPal(int num)
{
int cnt = 0;
for(int base=2;base<=10;++base){
if( isPal(num,base) ){
cnt++;
if(cnt==2)
return true;
}
}
return false;
}
void solve()
{
int n,s;
in>>n>>s;
while(n--){
while(!duaPal(++s));
out<<s<<endl;
}
}
int main(int argc,char *argv[])
{
solve();
return 0;
}
附原題：
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