HDOJ 1056 HangOver(水题)

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谙忆 发表于 2021/05/26 16:59:02 2021/05/26
【摘要】 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...

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)"); } }

}

  
 
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文章来源: chenhx.blog.csdn.net,作者:谙忆,版权归原作者所有,如需转载,请联系作者。

原文链接:chenhx.blog.csdn.net/article/details/51001305

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