poj1003

maksyuki 发表于 oj 分类，标签: 基础算法-简单数学, 数值-二分\三分\折半枚举

15 9月 2015

Hangover

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

题目类型：简单数学+二分

算法分析：先将所有的值预先处理出来，然后对于每一次的查询，二分答案

#include <cstdio>
#include <iostream>
#include <fstream>
using namespace std;
#define maxn 300
const double delta = 1e-8;

int zero (double x)
{
    if (x < -delta)
        return -1;
    return x > delta;
}

int main()
{
    double len[maxn];
    int i;
    len[0] = 0.0;

    for (i = 1; zero (len[i-1] - 5.20) < 0; i++)
        len[i] = len[i-1] + 1.0 / (i + 1);

    double x;
    //fstream cin ("aaa.txt");
    cin >> x;

    while (zero (x))
    {
        int left, right;
        left = 0, right = i;

        while (left + 1 < right)
        {
            int mid = (left + right) / 2;

            if (zero (len[mid] - x) < 0)
                left = mid;

            else
                right = mid;
        }
        printf ("%d card(s)\n", right);
        cin >> x;
    }
    return 0;
}

#include <cstdio>

#include <iostream>

#include <fstream>

using namespace std;

#define maxn 300

const double delta = 1e-8;

int zero (double x)

{

if (x < -delta)

return -1;

return x > delta;

}

int main()

{

double len[maxn];

int i;

len[0] = 0.0;

for (i = 1; zero (len[i-1] - 5.20) < 0; i++)

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

double x;

//fstream cin ("aaa.txt");

cin >> x;

while (zero (x))

{

int left, right;

left = 0, right = i;

while (left + 1 < right)

{

int mid = (left + right) / 2;

if (zero (len[mid] - x) < 0)

left = mid;

else

right = mid;

}

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

cin >> x;

}

return 0;

}

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