POJ 1003 Hangover （水题）

最新推荐文章于 2024-06-06 09:06:09 发布

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最新推荐文章于 2024-06-06 09:06:09 发布

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本文链接：https://blog.youkuaiyun.com/u013557725/article/details/38614417

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探讨如何通过堆叠卡片使其最大限度地悬空于桌面之外。输入一个数值，程序将计算并输出达到该悬空长度所需的最少卡片数量。

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Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 101382		Accepted: 49291

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.

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

1.00
3.71
0.04
5.19
0.00

Sample Output

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

题意：悬挂n张卡片的长度是x= 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) ；现在给你一个三位浮点数（0.01-5.20）问至少悬挂多少张卡片可以使卡片长度x>=给的浮点数，输入多样例，结束标志为0.00

题目链接：http://poj.org/problem?id=1003

代码：#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;


int main(){
    double len;
    while(cin>>len,len){
        int i=2;
        for(;;i++){
            len=len-1.0/i;
            if(len<=0) break;
        }
        cout<<i-1<<" card(s)"<<endl;
    }
    return 0;
}