POJ 百练 1003: Hangover

 

时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

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.

输出

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.

样例输入

1.00
3.71
0.04
5.19
0.00

样例输出

3 card(s)
61 card(s)
1 card(s)
273 card(s)

 

#include
int main(){
 float l,i,sum;
 while(scanf("%f",&l)!=EOF){
  if(l==0.00)
   break;
  sum=0;
  for(i=2.00;;i+=1.00){
   sum+=1.00/i;
   if(sum>=l)
    break;
  }
  printf("%d card(s)\n",(int)(i-1));
 }
 return 0;
}