POJ 1003 Hangover

Hangover

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 119902 Accepted: 58593

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

1.003.710.045.190.00

Sample Output

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

题目大意：给出一个数，求使得1/2+1/3+······+1/(n+1)>=c的最小的n。

思路：水题。题目上限5.20已给出，预处理一个数组a[]，用来存储S=1/2+1/3+······+1/(n+1)的值。判断和输出即可。

#include <iostream>#include<cstdio>using namespace std;int main(){    double a[300],x;    a[0]=0.00;    for(int i=1;;i++)    {        a[i]+=a[i-1]+1.00/(i+1);        if(a[i]>=5.20) break;    }    while(scanf("%lf",&x)!=EOF)    {        if(!x) break;        for(int i=1;;i++)        {            if(a[i]>=x) {printf("%d card(s)\n",i);break;}        }    }    return 0;}