NEFU OJ5 An interesting problem

題目：

                 有n位數d1d2d3...dn，如果將最後一位數dn移到首位，即dnd1d2d3...d(n-1)，且 dnd1d2d3...d(n-1) = dn*d1d2d3...dn，則稱這個數滿足條件。

                 輸入dn和k，求出尾數爲dn的滿足條件的第k個數。

                 目前只想到大數來做，但超時。暫時沒有更好方法，先放着

               

#include<iostream>
#include<string>
#include<stdio.h> 
using namespace std;

string bignumber = "1";
string lefts = "1";
int times,dn;
void addBig()
{
	int len = bignumber.length();
	for(int i =len-2;i>=0;--i)
	{
			int end = bignumber[i]-'0';
			++end;
			if(end>=10)
			{
				bignumber[i] = '0'+0;
				if(i == 0)
				{
					bignumber = lefts+bignumber;
					break;
				}
				else
					continue;
			}
			else
			{
				bignumber[i] = '0'+end;
				break;
			}
	}
}

bool judge()
{
	int len = bignumber.length();
	string temp = bignumber.substr(0,len-1);
	string templ ="s";
	templ[0] = bignumber[len-1];
	if(templ[0] != ('0'+dn) )
		return false;
	int begin = atoi(templ.c_str());
	templ += temp;
	int oldNum = atoi(bignumber.c_str());
	int newNum = atoi(templ.c_str());
	if(newNum == begin*oldNum)
	{
		--times;
		if(times == 0)
		{
			return true;
		}
	}
	return false;
} 
int main()
{
	while(cin>>dn>>times)
	{
		bignumber += '0'+dn;
		while(!judge())
		{
			addBig();
		}
		printf("%s\n",bignumber.c_str());
		bignumber.resize(1);
		bignumber = "1";
	}
	return 1;
}