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; //POJ1003 public 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。