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<iostream>
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;
}
}