poj 1003 Hangover

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 125133		Accepted: 61051

DescriptionHow 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 thebottom 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.InputThe input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c willcontain exactly three digits.OutputFor each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.Sample Input

1.00
3.71
0.04
5.19
0.00

Sample Output

3 card(s)
61 card(s)
1 card(s)
273 card(s)

<4000pre class="sio" style="font-family:'Courier New', Courier, monospace;font-size:14px;">题目解析：

这个题就是poj最水的题了，不用说什么了，直接看代码把，很容易理解的。

代码：

#include<iostream>using namespace std;int main(){double l,sum;int i,cnt;while(cin>>l&&l!=0.00){sum=0.0;cnt=0;for(i=2; sum<l; i++){sum+=1.0/i;cnt++;}cout<<cnt<<" "<<"card(s)"<<endl;}return 0;}