Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d
& %I64u
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
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.
Output
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.
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)
#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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