13、Hangover

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)

 

解题思路：刚开始看这道题，图文并茂还以为很难，结果是一道水题。不过既然做了就写一下：用了一个递归求n块板子的总悬挂长度，然后在主函数中比较输入长度和n块板子的总悬挂长度（n从1开始），当输入长度刚好小于或等于n块板子的悬挂长度时，按格式输出此时的n。

具体代码：

import java.util.*;
public class Main {
     public static float totalLength(int n){
          if(n==1)
              return (float)1/2;
          return (float)1/(n+1)+totalLength(n-1);
     }

     public static void main(String[] args) {
          Scanner scanner=new Scanner(System.in);
          float temp;
          int i;
          while((temp=scanner.nextFloat())!=0.00f){
              for(i=1;temp>totalLength(i);i++);
              System.out.println(i+" card(s)");
          }
     }
}

做题感悟：原来POJ上也有水题，就当保持手感。