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

[align=left]Problem Description[/align]
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.

 

[align=left]Sample Input[/align]

1.00
3.71
0.04
5.19
0.00

 

[align=left]Sample Output[/align]

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;
}