ZOJ Problem Set - 1045 HangOver

HangOver

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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.

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.

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.

Example input:

1.00
3.71
0.04
5.19
0.00

Example output:

3 card(s)
61 card(s)
1 card(s)
273 card(s)

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Source: Mid-Central USA 2001

分析：

套公式的水题，计算1/2+1/3+........+1/(n+1)<=c;

注意double的处理就行

ac代码：

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
    double c;
    int i;
    while(scanf("%lf",&c)&&c!=0.00)
    {
        double s=0;
        for(i=2;s<c;i++)
        {
            s+=1.00/i;//此处就是double型，1.00
        }
        printf("%d card(s)\n",i-2);
    }
}