using System;
using System.Collections.Generic;
using System.Text;
using System.Collections;
using System.IO;
 
namespace mathGame
{
class findTheNumber
{
static void Main(string[] args)
{
findTheNumber f = new findTheNumber();
f.showTheAnswer();
}
 
/**
*分析： 设目标为一个N位数，前N-1位为X，末位为Y
*       则，原数=10x+y；变换后=10^(N-1)*y+x；
*       (10x+y)*2=10^(N-1)*y+x
*       19x=(10^(N-1)-2)*y
*       由y为一个一位数，得到10^(N-1)-2为19的倍数，故有：
*
*/
ArrayList getN()
{
int n = 2;
findTheNumber f = new findTheNumber();
BigInteger a = f.mathPower(n-1)-2;
ArrayList al = new ArrayList();
do
{
if (a % 19 == 0)
{
al.Add(n);
 
}
 
n++;
a=f.mathPower(n-1)-2;
 
}while(a.dataLength<100);
return al;
}
 
BigInteger mathPower(int n)
{
BigInteger i ;
if (n == 0)
return 1;
else
{
return 10 * mathPower(n-1);
}
 
}
/**
*  通过getN()获取N值后，可得N=18，x=一个17位数*y;
*  满足题目的y值有2-9；
*
*/
void showTheAnswer()
{
BigInteger x;
int count = 0;
findTheNumber f = new findTheNumber();
ArrayList nlist = f.getN();
try
{
FileStream fs = new FileStream("numGame.txt", FileMode.OpenOrCreate);
StreamWriter sw = new StreamWriter(fs);
foreach (int n in nlist)
{
for (int y = 2; y <= 9; y++)
{
x = (f.mathPower(n - 1) - 2) * y / 19;
sw.WriteLine("Answer {0}:{1}", ++count, (10 * x + y));
}
}
sw.Close();
}
catch
{
Console.WriteLine("异常");
Console.ReadLine();
}
}
}
}