POJ - 1003 Hangover

[align=center]Hangover[/align]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 99582		Accepted: 48278

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
Mid-Central USA 2001

思路：sum = 1/2 + 1/3 + 1/4 + ...+ 1/(n+1)  ,直到sum >= c;

#include "iostream"
#include "cstdio"
#include "cstring"
#include "cmath"
#include "algorithm"
using namespace std;

int main()
{
//freopen("in.txt", "r", stdin);
int i, j;
float c;
while(scanf("%f", &c) == 1 && c != 0.00)
{
float sum = 0.00;
for(i = 2; i; i++)
{
sum += 1.0/i;
if(sum >= c)
{
cout<<i-1<<" card(s)"<<endl;
break;
}
}
}

return 0;
}