Hangover 分类： POJ 2015-06-11 10:34 12人阅读 评论(0) 收藏

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 108765		Accepted: 53009

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)
水题，不说了，还是看代码
#include <iostream>
#define exp 1e-6
using namespace std;
double sum[300];
int main()
{
    sum[0]=0;
    for(int i=1;i<=300;  i++)
    {
        sum[i]=(1.0/(i+1))+sum[i-1];
    }
    double len;
    while(cin>>len)
    {
        if(len<exp)
        {
            break;
        }
        for(int i=1;i<=300;i++)
        {
            if(len<sum[i])
            {
                cout<<i<<" card(s)"<<endl;
                  break;
            }

        }

    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。