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E - Hangover(1.4.1)

2024-07-11 21:59:14   221人阅读


Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u
Submit Status

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)
#include <iostream>     
#include<cmath>                
using namespace std;   
int main()
{
double  t[10000];
	t[0]=0;
	int i=0;
	for(;t[i]<=5.2;)
	{
		i++;
		t[i]=t[i-1]+1.00/(i+1);
	}
	double x;
	while(cin>>x&&x)
	{
		  int l, r;  
        l = 0;                  
        r = i  ;
        while (l + 1 < r) 
		{  
            int mid = (l + r) / 2;       
            if ((t[mid] - x)<-0.0000001)      
                l = mid;  
            else  
                r = mid;  
        }  
        cout << r << " card(s)" << endl;    
        
	}


return 0;
}


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