POJ-Hangover

Hangover
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 127984 Accepted: 62482

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

写题写着写着就会想能不能用新学的知识点，从而忘了最简单的处理方式。

附上代码：

#include<iostream>#include<cstdio>using namespace std;int main(){    double n;    while(scanf("%lf",&n)&&n!=0.00)    {        double sum=0;        for(int i=2;;i++)        {            sum=sum+(double)1/i;            if(sum>n||sum==n)            {                printf("%d",i-1);                break;            }        }        printf(" card(s)\n");    }    return 0;}