hdu-oj 1056 HangOver

HangOver

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

Total Submission(s): 9319    Accepted Submission(s): 3920

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

1.003.710.045.190.00

 

Sample Output

3 card(s)61 card(s)1 card(s)273 card(s)
题目大意：输出长度>=n所需要的木板数目；
注意：长度和应定义为double类型
附代码：
#include <stdio.h>int main(){   double sum,n;  int i;  while(scanf("%lf",&n)!=EOF&&n!=0)  {    sum=0;      for(i=1;;i++)      {  sum+=1.0/(i+1);        if(sum>=n)          break;      }      printf("%d card(s)\n",i);  }    return 0;}