1003

总时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

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.

输出

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.

样例输入

1.00
3.71
0.04
5.19
0.00

样例输出

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

分析：这道题相对简单。

 

 

解决方案一：G++

 

 

#include<iostream>
#include<cstdlib>
#include<cstdio>
using namespace std ;
int main()
{
     double ch ;
    while(cin>>ch){
    if(ch==0.00) break ;
     int i ;
     double t = 2.00 ;
     double sum = 0 ;
 for(i=0;ch>sum ;i++){
    sum += (double)1/t ;
    t += 1.00 ;
 }
 printf("%d card(s)\n",i) ;
   }
   return 0 ;
}

 

 或者

  

#include<iostream>
#include <cmath>
using namespace std;
int main(){
    float f;
    int i = 0 ;
    while(cin>>f)
   {
    if(f == 0.00)break;
    int n = 0;
    float sum = 0;
    float temp= -1;
    for(i = 2;f > sum;i++)
        {
            sum = sum+ (float)1/i;

    }
    cout<<i - 2<<" card(s)"<<endl;
    }

     return 0;
     }

 

 解决方案二：Java

 

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
       Scanner sc = new Scanner(System.in) ;      
       while(sc.hasNext())
       {   
    	   float t = sc.nextFloat() ;
    	   if(t==0.00) {break ; }
    	   int i ;
    	   float f = 2,sum=0 ;
    	   for(i=0;t>sum;i++)
    	   {
    		   sum += (float)1/f ;   
    		   f++ ;
    	   }
    	   System.out.println(i+" card(s)") ;
       }
	}

}

  或者

  

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
       Scanner sc = new Scanner(System.in) ;      
       while(sc.hasNext())
       {   
    	   float t = sc.nextFloat() ;
    	   if(t==0.00) {break ; }
    	   int i ;
    	   float sum=0 ;
    	   for(i=2;t>sum;i++)
    	   {
    		   sum += (float)1/i ;   
    		   
    	   }
    	   System.out.println(i-2+" card(s)") ;
       }
	}

}