POJ 1003 Hangover

探讨如何通过堆叠方式使卡片最大限度地悬空于桌面之外。利用数学原理计算不同数量卡片达到特定悬空长度的方法。

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Hangover
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 101308 Accepted: 49252

Description

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.



Input

The 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 will contain exactly three digits.

Output

For 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)

Source


#include <iostream>
using namespace std;
int main(){
	double a;
	while(cin>>a){
		if(a<0.0001)
			return 0;
		double num=0.0;
		for(int i=2;;i++){
			num+=1.0/i;
			if(num-a>0.0000001){
				cout<<i-1<<" card(s)"<<endl;
				break;
			}
		}
	}	
	return 0;
}


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