快速切题 poj 1003 hangover 数学观察 难度:0

Hangover

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 103896		Accepted: 50542

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)

题意:求sum(1/n)=给定数的n
思路:因为n只有可能300种,预处理比较即可,精度0.01,不需要设置eps

#include <iostream>

double tc;

double sum[300];

using namespace std;

int main(){

    ios::sync_with_stdio(false);

    double s=0;

    for(int i=0;i<300;i++){

        s+=1.00/(i+2);

        sum[i]=s;

    }

    while(cin>>tc&&tc){

        int i=0;

        for(int i=0;i<300;i++){

            if(sum[i]>=tc){

                cout<<i+1<<" card(s)"<<endl;

                break;

            }

        }

        if(i==300)cout<<"ERROR"<<endl;

    }

    return 0;

}