Section 1.2 dualpal

这道题与上面那道题不同，这道题要我们从S + 1开始找出 N个数，这N个数在2到10进制

中的形式至少有两个回文数。我们只需依次判断，符合要求就输出即可。

/*
ID:yucept21
LANG:C++
TASK:dualpal
*/

#include<cstdio>
#include<cstring>
#include<cstdlib>

const int MAXN = 50;
int N, S;
char s[MAXN];
char B[] = "0123456789" ;

bool ispal( char *s)
{
    int len = strlen( s);
    for( int i = 0; i <= len / 2; i ++)
        if( s[i] != s[len - 1 - i] )
            return false;
    return true;
}

int main()
{

    freopen( "dualpal.in", "r", stdin);
    freopen( "dualpal.out", "w", stdout);

    scanf( "%d%d", &N, &S);
    int cnt = 0;
    for( int i = S + 1; cnt != N; i ++)
    {
        int ans = 0;
        for( int j = 2; j <= 10; j ++)
        {
            int k = i;
            int top = 0;
            memset( s, 0, sizeof s);
            while( k != 0)
            {
                s[top ++] = B[k % j];
                k /= j;
            }
            bool flag = ispal( s);
            if( flag || top == 1) ans ++;
            if( ans == 2) {
                cnt ++;
                printf( "%d\n", i);
                break;
            }
        }
    }
    return 0;
}