URL: http://acm.zju.edu.cn/show_problem.php?pid=1045
Problem:
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.
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.
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.
Example input:
1.00
3.71
0.04
5.19
0.00
Example output:
3 card(s)
61 card(s)
1 card(s)
273 card(s)
My Solution:（C++，GCC）
I solve this problem in brute forced way. Although it has been accepted by the online-judge system, it is not an effective way due to my poor mathmatics knowledgement. There would another approach in which a formula will be used instead.
#include <iostream>
// PS: The code is for study purpose only, never submit it as ones own.
// From: http://blog.csdn.net/mskia/
// email: ichobits@21cn.com
int main(void) {
double len;
while (std::cin >> len && len != 0.00) {
double curlen = 0;
for (double i = 2; ; ++i) {
curlen += 1 / i;
if (curlen >= len) {
std::cout << i - 1 << " card(s)" << std::endl;
break;
}
}
}
return 0;
}