Hangover
来源:互联网 发布:成都曼巨网络 编辑:程序博客网 时间:2024/05/18 03:49
Hangover
- 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
1.003.710.045.190.00
- 样例输出
3 card(s)61 card(s)1 card(s)273 card(s)
-
#include <stdio.h>int main(){float c;while (scanf("%f", &c) && c){float sum = 0;for (int i = 2; i >= 2; i++){sum += 1.00 / i;if(sum >= c ){printf("%d card(s)\n", i-1);break;}}}return 0;}
- Hangover
- Hangover
- HangOver
- Hangover
- Hangover
- HangOver
- Hangover
- Hangover
- HangOver
- HangOver
- Hangover
- Hangover
- HangOver
- Hangover
- HangOver
- HangOver
- Hangover
- HangOver
- struct 级联下拉列表框
- JSTL----国际化标签库 fmt(转自网易)
- Android网络电话软件Sipdroid试用
- wpf 获取一张图片中的一部分
- 计算机中的数值数据表示
- Hangover
- 相关学习---链接
- StreamReader与StreamWriter乱码问题
- 返回最新添加的学生的自动编号Id
- 多个结果集的查询(使用table显示多个结果集)
- 小心C++的自动类型转换陷阱
- Ajax提交
- ADO.net小结
- NYOJ - A Famous Music Composer