一. 题目
Hangover
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 116593 | Accepted: 56886 |
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.
Input
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.
Output
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
二. 题意
- 按照固定的累加方式：从高层到低层，相对于其相邻下层可以伸出的长度为(1/2),(1/3),...,(1/n)
- 给定一个指定长度 S，问最少需要累加多上块板，使其相对于桌面的伸出长度大于或等于 S
三. 分析
- 算法核心: 此题比较简单，无需考虑任何算法
- 实现细节: 简单的浮点数累加运算即可
四. 题解
1 #include <stdio.h>
2
3 int main()
4 {
5 int i;
6 float length, sum;
7
8 while (1) {
9 sum = 0;
10 scanf("%f\n", &length);
11 if (!length) break;
12
13 for (i = 2; ; i++) {
14 sum += (1 / (float)i);
15
16 if (sum >= length) {
17 printf("%d card(s)\n", i - 1);
18 break;
19 }
20 }
21 }
22
23 return 0;
24 }