POJ1003-hangOver

       How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.

开始，尝试了很多算法，还感觉很难，后来发现直接暴力解决就可以，时间还是0MS,我果然想多了。

/*
*hang over 
*POJ 1003
*author:yuan tian
*whu eis
*先计算出0.00 - 0.52所有对应卡片值,进行储存
*/
#include
using namespace std;
int main()
{
	int cardIndex = 0;
	double queryNum = 0.0;
	double overHang = 0.0;
	while (cin >> queryNum)
	{
		overHang = 0.0;
		if (!queryNum)
			break;
		for (cardIndex = 1;; cardIndex++)
		{
			overHang += 1.0 / (double)(cardIndex + 1);
			if (overHang > queryNum)
			{
				break;
			}
		}
		cout << cardIndex << " card(s)" << endl;
	}
}