HDU 1056 HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8185    Accepted Submission(s): 3379

Problem 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.

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.
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)

 

Source

Mid-Central USA 2001

 

思路：高精度

 

代码：

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int i,j,k;
    double s[330];
    double sum = 0,n;
    j = 1;
    for(i = 2;i < 330;i ++)
    {
        sum = sum + (double)1 / i;
        k = (int)(sum * 100);
        s[j ++] = (double)k / 100;
    }
    while(~scanf("%lf",&n))
    {
        for(i = 1;i < 330;i ++)
              if(n >s[i - 1]&&n <= s[i])
              {
                      printf("%d card(s)
",i);
                      break;
              }
   }
}