每天一道算法_4_Hangover
此系列刚开始两天就被十一假期打断了，去山西玩了几天，今天刚回来，为了弥补一下心里的貌似隐隐作痛的愧疚感，挑了个简单的练练手，权当给自己补上一刀。
今天的题目是Hangover，如下：
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 makencards 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
Output
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)
代码如下：
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Hangover {
public static void main(String args[]){
List<Float> list = new ArrayList<Float>();
Scanner scanner = new Scanner(System.in);
int all = scanner.nextInt();
while(all > 0){
list.add(scanner.nextFloat());
all --;
}
for(int i = 0; i < list.size(); i ++){
float f = list.get(i);
int count = 1;
float j=2,sum=0;
for(; sum + (1/j) < f; j ++){
count ++;
sum = sum+ (1/j);
}
System.out.println(count + " card(s)");
}
}
}
输入输出如下：
4 1.00 3.71 0.04 5.19
3 card(s)
61 card(s)
1 card(s)
273 card(s)
由于输入的时候弄了半天没想出合适的用0.00结尾的方法，就用了开头先输入一个整数，表示后面要输入的整数个数，然后依次输入的方法，
相信你能看懂。
作者：jason0539
微博：http://weibo.com/2553717707
博客：http://blog.csdn.net/jason0539（转载请说明出处）