HDOJ 1056 HangOver（水题）

Problem 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.

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.

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)

英语不好真心看不懂（表示看不懂）。。。说了这么多，其实就是一个大水题。

公式题目给出了，就是求公式的和正好大于输入的n的时候的最小的整数。

import java.util.Scanner;
public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            double n = sc.nextDouble();
            if(n==0){
                return ;
            }
            int a=0;
            int t = 2;
            double sum=0;
            do{
                sum = sum+(1.0/t*1.0);
                a++;
                t++;
            }while(sum<n);
            System.out.println(a+" card(s)");
        }
    }
}