HDUOJ--1058HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7884    Accepted Submission(s): 3233

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

http://acm.hdu.edu.cn/showproblem.php?pid=1056

代码：

#include<stdio.h>
int main()
{
    double a,sum;
    int i;
    while(scanf("%lf",&a),a)
    {
        sum=0.0;
     for(i=2;i<=277;i++)
     {
         sum+=1.0/i;
       if(sum-a>=0) break;
     }
     printf("%d card(s)\n",i-1);
    }
    return 0;
}

数学题...就是搞不清要精确到哪一点..这样的，虽然 及其简单。。。但是往往AC率不高！！