HangOver
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7884 Accepted Submission(s): 3233
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 3 card(s) 61 card(s) 1 card(s) 273 card(s)
Source http://acm.hdu.edu.cn/showproblem.php?pid=1056 代码：
1 #include<stdio.h>
2 int main()
3 {
4 double a,sum;
5 int i;
6 while(scanf("%lf",&a),a)
7 {
8 sum=0.0;
9 for(i=2;i<=277;i++)
10 {
11 sum+=1.0/i;
12 if(sum-a>=0) break;
13 }
14 printf("%d card(s)\n",i-1);
15 }
16 return 0;
17 }
View Code
数学题...就是搞不清要精确到哪一点..这样的，虽然 及其简单。。。但是往往AC率不高！！
编程是一种快乐，享受代码带给我的乐趣！！！
