POJ 1003 Hangover

时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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 ncards 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.

输入

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.

输出

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.

样例输入

1.003.710.045.190.00

样例输出

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

 

（1）、源代码：

#include <iostream>

using namespace std;

 

int main(){

                int i;

                double length,sum;

                

                while(1){

                                cin >> length;

                                sum = 0;

                                if(length == 0.00)

                                                break;

                                for(i = 1; i < 300; i++){

                                                sum += 1.0/(i+1);

                                                if(sum >= length){

                                                                cout << i << " card(s)\n";

                                                                break;

                                                }

                                }

                }

                return 0;

}

  

（2）、解题思路：略

（3）、可能出错：略