zoukankan      html  css  js  c++  java
  • E


    Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

    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)
    #include <iostream>     
    #include<cmath>                
    using namespace std;   
    int main()
    {
    double  t[10000];
    	t[0]=0;
    	int i=0;
    	for(;t[i]<=5.2;)
    	{
    		i++;
    		t[i]=t[i-1]+1.00/(i+1);
    	}
    	double x;
    	while(cin>>x&&x)
    	{
    		  int l, r;  
            l = 0;                  
            r = i  ;
            while (l + 1 < r) 
    		{  
                int mid = (l + r) / 2;       
                if ((t[mid] - x)<-0.0000001)      
                    l = mid;  
                else  
                    r = mid;  
            }  
            cout << r << " card(s)" << endl;    
            
    	}
    
    
    return 0;
    }


  • 相关阅读:
    Microsoft 机器学习产品体系对比和介绍
    使用ANNdotNET进行情感分析
    使用.NET Hardware Intrinsics API加速机器学习场景
    关于ML.NET v0.6的发布说明
    强化学习的十大原则
    关于ML.NET v0.5的发布说明
    使用ML.NET实现基于RFM模型的客户价值分析
    使用ML.NET实现NBA得分预测
    Azure认知服务之Face API上手体验
    Orange——开源机器学习交互式数据分析工具
  • 原文地址:https://www.cnblogs.com/claireyuancy/p/6768938.html
Copyright © 2011-2022 走看看