HangOver

HangOver

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7693    Accepted Submission(s): 3129

Problem 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.

 

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.

 

 

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

 

题目没什么难度,分明就是某年NOIP的级数求和,不过题目里如果不说,我还真不一定能想到,这个结论得记一下.

#include<stdio.h>
#include<string.h>
int f[1024];
double s[1024];
void getprepared()
{
    memset(f,0,sizeof(f));
    memset(s,0,sizeof(s));
    s[1]=0.5;
    for (int i=2;i<=1000;i++) s[i]=s[i-1]+1.0/(i+1);
    for (int i=1;i<=520;i++)
    {
        double x=i/100.0;
        for (int j=1;j<=1000;j++)
        if (s[j]>=x)
        {
            f[i]=j;
            break;
        }
    }
}
int main()
{
    getprepared();
    double ss;
    while (scanf("%lf",&ss)!=EOF)
    {
        if (ss==0) return 0;
        int x=100*ss;
        printf("%d card(s)
",f[x]);
    }
    return 0;
}