[POJ] #1003# Hangover : 浮点数运算

一. 题目

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 116593		Accepted: 56886

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

二. 题意

• 按照固定的累加方式：从高层到低层，相对于其相邻下层可以伸出的长度为(1/2),(1/3),...,(1/n)
• 给定一个指定长度 S，问最少需要累加多上块板，使其相对于桌面的伸出长度大于或等于 S

三. 分析

• 算法核心: 此题比较简单，无需考虑任何算法
• 实现细节: 简单的浮点数累加运算即可

四. 题解

 #include <stdio.h>

 int main()
 {
     int i;
     float length, sum;

     while () {
         sum = ;
         scanf("%f\n", &length);
         if (!length) break;

         for (i = ; ; i++) {
             sum += ( / (float)i);

             if (sum >= length) {
                 printf("%d card(s)\n", i - );
                 break;
             }
         }
     }

     return ;
 }