OpenJudge / Poj 1003 Hangover

链接地址：

Poj：http://poj.org/problem?id=1003

OpenJudge：http://bailian.openjudge.cn/practice/1003

题目：

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 94993		Accepted: 46025

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：

水题

代码：

 #include "stdio.h"
 //#include "stdlib.h"
 int main()
 {
     float a,sum;
     int i;
     scanf("%f",&a);
     while(a!=)
     {
         i = ;
         sum = ;
         while(sum < a)
         {
             sum += ((float)/i);
             i++;
         }
         printf("%d card(s)\n",i-);
         scanf("%f",&a);
     }
     //system("pause");
     return ;
 }