[UVA][重心] 651 - Deck
Deck
| Deck |
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
The input file 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
1 2 3 4 30
Sample Output 1
12345678901234567
# Cards Overhang
1 0.500
2 0.750
3 0.917
4 1.042
30 1.997
Footnotes
- ... Output1
- 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.
Miguel A. Revilla
2000-01-17
參照解法 http://jackchen0227.iteye.com/blog/1104352
要明白相同的疊法會累積在上一層, 而最下層會改變。
使用遞推的方式去想
#include <stdio.h>
int main() {
int n, i;
puts("# Cards Overhang");
while(scanf("%d", &n) == 1) {
double ret = 0;
for(i = 1; i <= n; i++)
ret += 1.0/2/i;
printf("%5d %.3lf\n", n, ret);
}
return 0;
}
/*
从上面的卡片往下看，设前n－1个超越的长度是L(n-1),则再加上最下面的一个共n个超越的长度设为L(n)，
可将上面的n-1个看成一个整体，最下面的单独看，
则由力矩平衡可得，
(n-1)m×(L(n)-L(n-1))=m×[1/2-(L(n)-L(n-1))]
（其中m为各个卡片的质量）,化简下即得递推关系式：L(n)-L(n-1)=1/2n
http://jackchen0227.iteye.com/blog/1104352
*/