杭电1056--Hangover(打表)

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10631    Accepted Submission(s): 4537

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

1 3 card(s) 

2 61 card(s) 

3 1 card(s) 

4 273 card(s)

 

 

 

Source

Mid-Central USA 2001

 

 

//打表找出范围，再遍历；double对应“lf”;

 1 #include<stdio.h>

 2 double sieve[300];

 3 int main()

 4 {

 5     int i; double sum=0.0;

 6     for(i=1;i<277;i++)

 7     {

 8         sum+=1.0/(i+1);

 9         sieve[i]=sum;

10     }

11     double length;

12     while(~scanf("%lf",&length),length)

13     {

14         for(i=1;i<277;i++)

15         {

16             if(sieve[i]>=length)

17             {

18                 printf("%d card(s)\n",i);

19                 break;

20             }

21         }

22     }

23     return 0;

24 }