题目：
HangOver
Time Limit: 1 Second Memory Limit: 32768 KB
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.
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.
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.
Example input:
1.00
3.71
0.04
5.19
0.00
Example output:
3 card(s)
61 card(s)
1 card(s)
273 card(s)
Source: Mid-Central USA 2001
题目大意：你可以把一叠卡片放的离桌子多远？如果有一张卡片，那么可达到的最远距离是卡片唱的的一半（假设卡片必须与桌子的边缘垂直），是用两张卡片，是上面一张能放到的最远距离超过下面长度的一半，而下面一张超过桌子的是卡片长度的1/3，所以能达到的最远距离是：1/2+1/3=5/6。一半来说，n张卡片能达到的最远距离是1/2+1/3+1/4+....+1/(n+1)，也就是最顶上的卡片超出第二张的1/2，第二张超出第三张的1/3，第三张超出第四张的1/4，等等，输入一行数字0.00时表示输入结束，每个测试例一行，是一个浮点数c（0.01<=c<=5.20），c刚好是3位数字。对每个测试例，输出达到距离c所需要的最少卡片的数量。注意输出格式！
算法分析：其实这道题就是1/2+1/3+...+1/(n+1)<=c，利用循环语句知道上面的等式超过c为止。
代码：
语言：c
#include<stdio.h>
int main()
{
float c;
while(scanf("%f",&c)&&c!=0)
{
int i=2;
float result=0;
while(result<c)
{
result=result+1.00/i;
++i;
}
printf("%d card(s)/n",i-2)
}
return 0;
}