Poj1003 Hangover, poj1003hangover
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 makeNCards 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 already strated 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
1.003.710.045.190.00
Sample Output
3 card(s)61 card(s)1 card(s)273 card(s)
import java.util.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); String str; int maxCardNum = 0; while(cin.hasNext()) { str = cin.nextLine(); if(str.equals("0.00")) break; float len = Float.valueOf(str).floatValue(); maxCardNum = getCardNum(len); System.out.println(maxCardNum + " card(s)"); } } private static int getCardNum(float len) { float value = 0; int index = 2; while (value < len) { value += 1.0/index; index++; } return index-2; } }
ACM Hangover (Peking University ACM1003) teaches C Language
Double format control should use % lf instead of % ld
POJ 1003, simple code error correction
Poj.org/problem? Id = 1003
Hangover
Not this question.