ACM pku 1003 解题报告(练习输入输出)

最新推荐文章于 2019-01-14 22:24:27 发布

manio

最新推荐文章于 2019-01-14 22:24:27 发布

阅读量2.2k

点赞数

分类专栏： Algorithm 文章标签： output input float table each 算法

本文链接：https://blog.youkuaiyun.com/manio/article/details/584317

版权

Algorithm 专栏收录该内容

33 篇文章

订阅专栏

探讨如何利用特定数量的卡片实现最大化的悬空距离，通过数学计算得出不同悬空长度所需的最少卡片数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Hangover
Time Limit:1000MS Memory Limit:10000K
Total Submit:12059 Accepted:4808

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
Mid-Central USA 2001

代码如下,效率不是很高,用了15MS,一次性通过了.算法优化以后再讲了.

main()
{
float min[5000];
int num[5000];
int i,total;
float temp;

scanf("%f",&min[0]);
i = 0;

while(min[i]){
i++;
scanf("%f",&min[i]);

}
total = i; /*total是输入的个数*/

/*求每个输入的不小于MIN[]的板数*/
for( i = 0 ; i < total ; i++ ){
temp = 0;
num[i] = 2;

while( temp < min[i] ){
temp =temp + ( 1.0 / num[i] ) ; /*注意这个地方一定要用1.0哦,要不NUM[I]也被隐式转化为整型了*/
num[i]++;