E - Hangover(1.4.1)

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

Submit 

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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;
}