HDU1330_Deck【水题】-CFANZ编程社区

Deck

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1206 Accepted Submission(s): 698

Problem Description

A single playing card can be placed on a table, carefully, so that the short edges of the card are parallel to the table's edge, and half the length of the card hangs over the edge of the table. If the card hung any further out, with its center of gravity off the table, it would fall off the table and flutter to the floor. The same reasoning applies if the card were placed on another card, rather than on a table.

Two playing cards can be arranged, carefully, with short edges parallel to table edges, to extend 3/4 of a card length beyond the edge of the table. The top card hangs half a card length past the edge of the bottom card. The bottom card hangs with only 1/4 of its length past the table's edge. The center of gravity of the two cards combined lies just over the edge of the table.

Three playing cards can be arranged, with short edges parallel to table edges, and each card touching at most one other card, to extend 11/12 of a card length beyond the edge of the table. The top two cards extend 3/4 of a card length beyond the edge of the bottom card, and the bottom card extends only 1/6 over the table's edge; the center of gravity of the three cards lines over the edges of the table.

If you keep stacking cards so that the edges are aligned and every card has at most one card above it and one below it, how far out can 4 cards extend over the table's edge? Or 52 cards? Or 1000 cards? Or 99999?

Input

Input contains several nonnegative integers, one to a line. No integer exceeds 99999.

Output

The standard output will contain, on successful completion of the program, a heading:

# Cards Overhang

(that's two spaces between the words) and, following, a line for each input integer giving the length of the longest overhang achievable with the given number of cards, measured in cardlengths, and rounded to the nearest thousandth. The length must be expressed with at least one digit before the decimal point and exactly three digits after it. The number of cards is right-justified in column 5, and the decimal points for the lengths lie in column 12.

Sample Input

Sample Output

The line of digits is intended to guide you in proper output alignment, and is not part of the output that your solution should produce.

12345678901234567

# Cards Overhang

1 0.500

2 0.750

3 0.917

4 1.042

30 1.997

题目大意：在桌子边上叠纸牌，使纸牌超出桌边的长度最长，并且不能掉下去。

即重心最多在桌子边缘上，问给你N张纸牌，最长能超出桌子边缘多长。

思路：这道题和POJ1607是一样的，需要用到物理学力矩的原理，作为一个物理

不好的弱渣来说真是无解啊。

参考博客：

http://gd2011.teacher.com.cn/UserLog/UserLogComment.aspx?UserlogID=15319

把第一块木板的重心放在第二块木板的右边缘，把这两块木板的重心放在第三块

木板的右边缘，把这三块木板的重心放在第四块木板的右边缘⋯⋯利用杠杆原理

可以推出，如果每块木板都是单位长，那么 n 块木板可以伸出桌面 (1 + 1/2 +

1/3 + … + 1/n) / 2 个单位的长度。

最终得到一个递推式：a[i] = a[i-1] + 1.0/i/2

#include<stdio.h>
#include<string.h>

int main()
{
    double a[100010];
    a[1] = 0.5;
    int N;
    for(int i = 2; i <= 100000; i++)
        a[i] = a[i-1] + 1.0/i/2;
    printf("# Cards  Overhang\n");
    while(~scanf("%d",&N))
    {
        printf("%5d%10.3lf\n",N,a[N]);
    }

    return 0;
}