学步园

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

1.00
3.71
0.04
5.19
0.00

样例输出

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

#include <stdio.h>
#include <math.h>
int main()
{
  float i;
  while(scanf("%f",&i)==1)
  {
    if(i<0.01 || i>5.20 )
    {   return 0;   }
    float tmp    = 0;
    float length = 0;
    int   number = 1;
    while(tmp < i)
    {
      length = 1.0/++number;
      tmp += length;
    }
    int card = number - 1;
    printf("%d card(s)\n",  card);
  }

  return 0;
}

另外，也可有数学解法，因为这个一个调和级数，所有有

其中γ是欧拉-马歇罗尼常数，而约等于，并且随着k 趋于正无穷而趋于0。