PKU ACM 1016 Numbers That Count

题目链接：Numbers That Count

主要思路：

这题不涉及什么算法，只是简单的字符串处理。需要注意的地方就是最终答案的优先级和边界的处理。

一开始我用了 _itoa 函数，一直WA，很奇怪。后来自己动手写了整形到字符串的函数，终于才AC了。

看来有时候还是自己动手，丰衣足食啊

 

源代码：

#include 
#include 
using namespace std;

void inttostr(int i, string& outstr)
{
	int number = 0;
	do
	{
		number = i % 10;
		outstr = (char)(number + '0') + outstr;
		i = i / 10;
	}while(i);
}

string GetInventory(string s)
{
	string inventory = "";
	string sn = "";
	string si = "";

	int i = 0;
	int board[10] = {0};
	for(i = 0; i < s.length(); i++)
	{
		board[s[i] - '0']++;
	}

	for(i = 0; i <= 9; i++)
	{
		sn = "";
		si = "";
		if(board[i] == 0)	continue;
		inttostr(board[i], sn);
		inttostr(i, si);
		inventory = inventory + sn + si;
	}

	return inventory;
}

int main()
{
	string result[17];
	int times;
	int i;

	while(true)
	{
		for(i = 0; i < 17; i++)
		{
			result[i] = "";
		}

		cin >> result[0];
		if(result[0] == "-1")	break;

		times = 1;
		while(times)
		{
			if(times > 15)
			{
				cout << result[0] << " can not be classified after 15 iterations " << endl;
				break;
			}

			result[times] = GetInventory(result[times - 1]);
			if(result[1] == result[0])
			{
				cout << result[0] << " is self-inventorying " << endl;
				break;
			}

			if(result[times] == result[times - 1])
			{
				cout << result[0] << " is self-inventorying after " << times - 1 << " steps" << endl; 
				break;
			}

			for(i = 0; i < times; i++)
			{
				if(result[i] == result[times])
				{
					cout << result[0] << " enters an inventory loop of length " << times - i << endl;
					break;
				}
			}

			if(i < times)	break;
			times++;
		}
	}
	
	return 0;
}