poj 1003

如题：http://poj.org/problem?id=1003



Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 107427		Accepted: 52397

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)

Source
Mid-Central USA 2001




思路：先打个表。然后再表里找一个符合项。

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

double len;

double card[500];

void init()

{

int i;

for(i=1;i<=500;i++)

card[i]=card[i-1]+(double)1.0/(i+1);

}

int main()

{

// freopen("C:\\1.txt","r",stdin);

init();

while(~scanf("%lf",&len))

{

if(len*100==0)

break;

int i;

for(i=1;i<500;i++)

if(card[i]>=len)

break;

printf("%d card(s)\n",i);

}

return 0;

}