poj-1003-Hangover

poj-1003-Hangover

作者语

深知自己的代码能力太差,开始在poj水第一题

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)

AC代码

#include <iostream>
#include <stdio.h>

using namespace std;

int main()
{
    double n;
    while(cin>>n,n)
    {
        double sum = 0;
        int x;
        for(double i=2;sum<n;i++)//如果sum>=n时满足题意
        {
        
            sum=sum+1.0/i;
            x=i;
            
        }
        cout<<x-1<<" card(s)"<<endl;
    }
}

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