USACO SEC.1.2 No.5 Dual Palindromes

题意：输入一个N和S，在[S + 1, 10000]范围以内找到N个至少在[2,10]两种进制上面都是回文数的数字

核心：仍然是转换十进制到任意进制的算法

注意G++下要包含<cstring>才能找到strcpy等相关的函数

/*
ID: lsswxr1
PROG: dualpal
LANG: C++
*/
#include <iostream>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <fstream>
using namespace std;

///宏定义
const int  INF = 1000000000;
const int MAXN = 15;
const int maxn = MAXN;
///全局变量 和 函数

#define USACO
#ifdef USACO
#define cin fin
#define cout fout
#endif
//////////////////////////////////////////////////////////////////////////
int N, S;
char str[maxn];
void numtrans(char* tar, int jingzhi, int num)
{
    if (num == 0)
    {
        strcpy(tar, "");
        return;
    }
    numtrans(tar, jingzhi, num / jingzhi);
    int len = strlen(tar);
    tar[len] = '0' + num % jingzhi;
    tar[len + 1] = '\0';
}
bool isPalindromes(char* src)
{
    int len = strlen(src);
    for (int i = 0; i < len / 2; i++)
    {
        if (src[i] != src[len - 1 - i])
            return false;
    }
    return true;
}
int main()
{

#ifdef USACO    
    ofstream fout ("dualpal.out");
    ifstream fin ("dualpal.in");
#endif    
    ///变量定义
    while (cin >> N >> S)
    {
        int cnt = 0;
        int cnttrans;
        for (int i = S + 1; S < 10000; i++)
        {
            cnttrans = 0;
            for (int j = 2; j <= 10; j++)
            {
                numtrans(str, j, i);
                if (isPalindromes(str))
                {
                    cnttrans++;
                    if (cnttrans >= 2)
                        break;
                }
            }
            if (cnttrans >= 2)
            {
                cnt++;
                cout << i << endl;
                if (cnt == N)
                {
                    break;;
                }
            }
        }
    }

    ///结束
    return 0;
}