FOJ_1002_HangOver

 

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)

通过这题我总算明白了,float已经可以被xx了...

精度问题WA了N次T.T

那时候不懂还老问老师....

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