POJ1003 Hangover

问题描述

总时间限制: 1000ms 内存限制: 65536kB
描述

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.

输入

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.

输出

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.
样例输入

1.00
3.71
0.04
5.19
0.00

样例输出

3 card(s)
61 card(s)
1 card(s)
273 card(s)

import java.util.ArrayList;import java.util.Scanner;//POJ1003public class Main{    static double[] arr=new double[280];    public static void main(String[] args) {        // TODO Auto-generated method stub        Scanner scanner=new Scanner(System.in);         arr[0]=0;        for(int i=2;i<280;i++){            arr[i-1] = arr[i-2]+(double)1/i;            //System.out.println(arr[i-1]);        }        double d;        while(true){            d=scanner.nextDouble();            if(d==0.00)                break;            cal(d);        }    }    private static void cal(double d) {        int i=0;        while(d > arr[i]){            i++;        }

小结：本题实质就是给出一个数字d(0.00<d<5.20),求出n，使得 1/2+1/3+1/4+……1/n<d<1/2+1/3+……+1/n+1/(n+1).输入数据以0.00作为结束标识。