HDU 1056 HangOver(數學題)

                                                                       HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11757    Accepted Submission(s): 5102

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

1.003.71

0.04

5.19

0.00

Sample Output

3 card(s)61 card(s)

1 card(s)

273 card(s)

題意：一直 1/2 

+

 1/3 

+

 1/4 

+

 ... 

+

 1/(

n

+

 1)疊加，直到>m. （和poj一模一樣的題）

AC代碼：

#include<cstdio>
int main()
{
  float a=0.0;
  while(1)
  {
      scanf("%f",&a);
    float sum=0;
    float i=0;
    int count=0;
    if(a==0.00)
    {
      break;
    }
    for(i=2.0;sum<a;i++)
    {
      sum=sum+1/i;
      count++;
    }
    printf("%d card(s)\n",count);
  }
  return 0;
}