ZOJ Problem Set – 1045 HangOver

     真的是不行了，很长时间没有玩过这玩意儿了，一下子上来，还真是很不习惯。碰了很多壁，终于碰到了这么一道相对比较简单的题目，也算是捡回一点点自信吧，希望以后会越走越好。

      下面先给出原题：

ZOJ Problem Set - 1045

HangOver

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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.

Example input:
1.00
3.71
0.04
5.19
0.00

Example output:
3 card(s)
61 card(s)
1 card(s)
273 card(s)

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Source: Mid-Central USA 2001

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=45

题目不难，直接给出代码吧：

  1: #include<iostream>
  2: #include<vector>
  3: 
  4: using namespace std;
  5: 
  6: const double MINVALUE = 0.00000001;
  7: void IntialPro(vector<double>& vec)
  8: {
  9:     double hangLength = 0.0;
 10:     for(int n = 1;hangLength <= 5.20; n++)
 11:     {
 12:         hangLength += 1/static_cast<double>(n + 1);
 13:         vec.push_back(hangLength);
 14:     }
 15: }
 16: int main(void)
 17: {
 18:     vector<double> hangOver;
 19:     IntialPro(hangOver);
 20: 
 21:     double input;
 22: 
 23:     while(cin>>input)
 24:     {
 25:         if(input > -1*MINVALUE && input < MINVALUE)
 26:         {
 27:             break;
 28:         }
 29:         for(size_t i = 0; i < hangOver.size(); i++)
 30:         {
 31:             if(hangOver[i] > input)
 32:             {
 33:                 cout<<i + 1<<" card(s)"<<endl;
 34:                 break;
 35:             }
 36:         }
 37:     }
 38:     return 0;
 39: }

在这段代码中，没有什么特别的需要主要的，唯一让我感觉到有点意思的是double类型与0.00进行比较的时候的特点。众所周知，double类型在内部存储的时候是存在精确缺失的，所以，要让double与0进行比较的时候，不能直接来，需要定义一个很小的值，就像我程序中的那样进行比较。这种技巧，我也是在某次朋友说要考考我的时候得知的。也许在验证条件不是很苛刻的时候，double直接与0比较的时候也能通过，但是一旦比较严格的话，无疑会失败的。

       这同时，让我想到了其它的几个比较。也是编程习惯的问题（在《高质量C、C++编程指南》中有讲）。如下：
bool值的比较(其中flag是一个bool类型的变量)

   if(flag)

指针类型与零的比较(其中p的定义为：char *p;)：

   if(p == NULL)