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 +
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) 法一：普通搜索法：
代码：
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int i;
double n,sum;
while(cin>>n&&n!=0.00)
{
sum=0.00;
for(i=1;i<350;i++)
{
sum=sum+1.0/double(i+1);
if(sum>=n) break;
}
cout<<i<<" card(s)"<<endl;
}
return 0;
}法二：二分查找
代码：
#include <iostream>
using namespace std;
const double delta=1e-8;
const int maxn=300;
int zero(double x)
{
if(x <-delta)
return -1;
return x>delta;
}
int main()
{
double len[maxn];
int total;
len[0]=0.0;
for(total=1;zero(len[total-1]-5.20)<0;total++)
{
len[total]=len[total-1]+1.0/double(total+1);
}
double n;
cin>>n;
while(zero(n))
{
int l,r;
l=0;
r=total;
while(l+1<r)
{
int mid=(l+r)/2;
if(zero(len[mid]-n)<0)
l=mid;
else
r=mid;
}
cout<<r<<" card(s)"<<endl;
cin>>n;
}
return 0;
}
