E – Hangover(1.4.1)
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
contain exactly three digits.
Output
Sample Input
|
1 2 3 4 5 |
1.00 3.71 0.04 5.19 0.00 |
Sample Output
|
1 2 3 4 |
3 card(s) 61 card(s) 1 card(s) 273 card(s) |
|
1 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 |
#include <iostream> #include<cmath> using namespace std; int main() { double t[10000]; t[0]=0; int i=0; for(;t[i]<=5.2;) { i++; t[i]=t[i-1]+1.00/(i+1); } double x; while(cin>>x&&x) { int l, r; l = 0; r = i ; while (l + 1 < r) { int mid = (l + r) / 2; if ((t[mid] - x)<-0.0000001) l = mid; else r = mid; } cout << r << " card(s)" << endl; } return 0; } |
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